qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,584,927 | <p>If I have two isomorphic groups, can I write <span class="math-container">$A \xrightarrow{\sim} B$</span> rather than <span class="math-container">$A \cong B$</span> to mean "A is isomorphic to B", or is the arrow notation only used if I have a map <span class="math-container">$\varphi : A \xrightarrow{\sim} B$</span> ?</p>
| Shaun | 104,041 | <p>Both notations are accepted. </p>
<p>If you're using either of them, though, it is good practise to state what it means; it is usually left as implicit from the context.</p>
<p>Having said that, it should be noted that "<span class="math-container">$A\stackrel{\sim}{\to}B$</span>" usually refers to a specific isomorphism (from <span class="math-container">$A$</span> to <span class="math-container">$B$</span>).</p>
|
2,202,048 | <p>There's a box with $N$ balls in it. One of them is red and the others white. What's the probability of getting the red ball at the $k$th try ( if you're not putting them back in?) where $k = 1,2,3,\ldots,N$</p>
| Fimpellizzeri | 173,410 | <p>You can compute the probability to be $1/N$, independently of $k$, directly by noticing that for the first $k-1$ draws you mustn't draw the red ball.</p>
<p>That said, another way to see this is that drawing a single ball from the box in succession until it is empty can be represented by an ordering of the balls (where the $i$-th ball represents the $i$-th draw). In that case, I believe it's pretty clear that the red ball is equally likely to show up in any two positions, and hence the probability does not depend on $k$.</p>
|
3,178,648 | <blockquote>
<p>We assign to every element <span class="math-container">$i$</span> from <span class="math-container">$N=\{1,2,...,n\}$</span> a positive integer <span class="math-container">$a_i$</span>. Suppose <span class="math-container">$$a_1+a_2+...+a_n = 2n-2$$</span> then prove that map <span class="math-container">$T: \mathcal{P}(N) \to \{1,2,...,2n-2\}$</span> defined with <span class="math-container">$$T(X) = \sum _{i\in X}a_i$$</span> is surjective. </p>
</blockquote>
<hr>
<p>We can assume that <span class="math-container">$a_1\leq a_2\leq ...\leq a_n$</span>. </p>
<p>Clearly, <span class="math-container">$a_1 = a_2 = 1$</span> and thus <span class="math-container">$1,2,2n-3,2n-4$</span> are in a range. </p>
<p>Also, if <span class="math-container">$a_i=2$</span> for some <span class="math-container">$i$</span> then we could easily apply induction. </p>
<p>Say <span class="math-container">$b_1< b_2<...<b_k$</span> are all different values that appear among <span class="math-container">$a_i$</span>. </p>
<p>Then we have <span class="math-container">$n _1\cdot b_1+n_2\cdot b_2+...+n_k \cdot b_k = 2n-2$</span> and <span class="math-container">$n_1+n_2+..+n_k = n$</span>. We have to prove that for each <span class="math-container">$l\leq 2n-2$</span> we have <span class="math-container">$$n' _1\cdot b_1+n'_2\cdot b_2+...+n'_k \cdot b_k = l$$</span></p>
<p>for some <span class="math-container">$n'_i\leq n_i$</span>. And here it stops. I have no idea how to find all those <span class="math-container">$n_i'$</span>. Any ideas?</p>
| antkam | 546,005 | <p>All the answers are basically saying the same thing, so I will simply try to say it in the most "graph-theoretic" way. :)</p>
<hr>
<p>Construct an <span class="math-container">$n$</span>-node tree <span class="math-container">$G$</span> with the node degrees being <span class="math-container">$a_1, a_2, ..., a_n$</span>. [This can be done, e.g. by starting with the highest numbers and keep combining them.]
Initially every node is colored black.</p>
<p>We will be coloring some nodes red, and the red nodes will represent <span class="math-container">$X \in \mathcal{P}(N)$</span>. Thus <span class="math-container">$T(X) = $</span> sum of node degrees of the red nodes. </p>
<p>Initially, every node is black, i.e. <span class="math-container">$X = \emptyset, T(X) = 0$</span>. We now increment <span class="math-container">$T(X)$</span> one by one.</p>
<ul>
<li><p>If some leaf is black, color it red. This increments <span class="math-container">$T(X)$</span> by <span class="math-container">$1$</span>.</p></li>
<li><p>If all leaves are red, pick <em>any black internal node</em> <span class="math-container">$v \in G$</span>. Suppose its degree is <span class="math-container">$a > 1$</span>. Since <span class="math-container">$G$</span> is a tree, <span class="math-container">$G - \{v\}$</span> is a new graph which is a collection of <span class="math-container">$a$</span> separate trees (including possibly singleton nodes). Each of these <span class="math-container">$a$</span> smaller trees has at least one leaf belonging to the original <span class="math-container">$G$</span>. Thus the original <span class="math-container">$G$</span> has at least <span class="math-container">$a$</span> leaves (which are, by assumption, all red). Now, change <span class="math-container">$v$</span> from black to red, and change any <span class="math-container">$a-1$</span> leaves from red back to black. This increments <span class="math-container">$T(X)$</span> by <span class="math-container">$1$</span>.</p></li>
</ul>
<p>We can keep doing this until all nodes are red, at which point <span class="math-container">$T(X) = 2n-2$</span>. Thus <span class="math-container">$T()$</span> goes through <span class="math-container">$\{0, 1, 2, ..., 2n-2\}$</span>, proving it is surjective.</p>
|
2,912,401 | <p>I'm trying to prove that the sequence $\left(\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\cdots\right)$ is not a Cauchy sequence. </p>
<p>I know that a sequence of real numbers is <em>not</em> Cauchy if there exists an $\epsilon>0$ such that, for all $N\in\mathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|\geq\epsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $\cdots\frac{99}{101},\frac{100}{101},\frac{1}{102},\cdots$) keeps growing larger. However, I'm not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|\geq\epsilon$. Thanks in advance for any help!</p>
| Mark Bennet | 2,906 | <p>You only need such an $\epsilon$ to exist, so you can choose a convenient value.</p>
<p>Then you need to show that there are gaps bigger than $\epsilon$ between elements of the sequence as far a=out as you care to go - that gaps as big as $\epsilon$ don't fade out and disappear in the tail of the sequence.</p>
<p>Well you have identified some chunky gaps which persist (you don't need every gap to be big). What $\epsilon$ would work for the gaps you have identified?</p>
|
3,122,612 | <p>I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:</p>
<p><span class="math-container">$\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n$</span></p>
<p>I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).</p>
<p><a href="http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf" rel="nofollow noreferrer">http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf</a></p>
<p>EDIT: The full equation is</p>
<p><span class="math-container">$\frac{\partial A}{\partial z}+\sum_{n=1}^{\infty}\frac{(-i)^{n-1}}{n!}k^n \frac{\partial^{n} A}{\partial t^{n}}=-i \frac{\chi^{(2)}\omega}{2nc}AA^{*}e^{-i\Delta \textbf{k}\cdot \textbf{z}}$</span></p>
<p>so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..</p>
| JV.Stalker | 416,274 | <p>Think the name of the function is "Exponential generating function (EGF)" see Wikipedia. </p>
<p><span class="math-container">$\sum\limits_{n=1}^\infty ik_n \frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$</span></p>
|
1,025,588 | <p>I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem:</p>
<p>If $a, b, c \gt 0$ prove that $$ a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.$$</p>
<p>I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you!</p>
| Community | -1 | <p>$$\color{blue}{a^2(a-b)+b^2(b-c)+c^2(c-a)=(a^2-c^2)(a-b)+(b^2-c^2)(b-c)}$$</p>
<p>$$\color{red}{a\geq b \geq c \implies (a^2-c^2)(a-b)+(b^2-c^2)(b-c) \geq0}$$ </p>
|
2,174,454 | <p>My teacher substitutes for this $\sum_{x=0}^{y} ({{y!} \over {x! (y-x)!} })$ by $2^y$, so I tried to use the (Mathematical induction) to prove it (My teacher did not ask me to do that, however I want to do it only to make sure for this statement)</p>
<p>My attempt:</p>
<ul>
<li>when $y=1$ :</li>
</ul>
<p>L.H.S.
$2^y=2$</p>
<p>R.H.S.
$\sum_{x=0}^{1} {{y!} \over {x! (y-x)!} } =2$</p>
<p>So this true when $y=1$</p>
<ul>
<li>Let the statement true when $y=k$ , so </li>
</ul>
<p>$\sum_{x=0}^{k} {{k!} \over {x! (k-x)!} } =2^k$</p>
<ul>
<li>When $y=k+1$</li>
</ul>
<p>L.H.S.</p>
<p>$2^y=2^{k+1}=2*2^k=2 * \sum_{x=0}^{k} {{k!} \over {x! (k-x)!} }$</p>
<p>R.H.S</p>
<p>$\sum_{x=0}^{k+1} {{(k+1)!} \over {x! (k-x+1)!} }=\sum_{x=0}^{k+1} {({k+1)k!} \over {x! (k-x+1)(k-x)!} }$</p>
<p>But now I don't know how can I complete it ? </p>
| MAN-MADE | 422,850 | <p>You can verify it easily. The terms in LHS are $\frac{y!}{x!(y-x)!}$ which means the way you can choose $x$ objects form $y$ objects. The sum over $x$ will give you total ways to select a object from $y$ objects which is $2^y$ since it is about select or not select i.e. 2 ways, i.e. in all total $2^y$ ways.</p>
<p>$(1+p)^y= \sum_{x=0}^{y} \frac{y!}{x!(y-x)!} p^x$, take $p=1$</p>
|
14,527 | <p>The <a href="https://en.wikipedia.org/wiki/Robinson%E2%80%93Schensted%E2%80%93Knuth_algorithm" rel="nofollow noreferrer">Robinson-Schensted correspondence</a> is a bijection between elements of the symmetric group and ordered pairs of standard tableaux of the same shape.</p>
<p>Some simple operations on tableaux correspond to simple operations on the group: switching the tableaux corresponds to inverse on the group.</p>
<blockquote>
<p>What about taking the transpose of the tableaux? Does that correspond to something easily described on permutations?</p>
</blockquote>
<p>In order to rule out easy guesses, let me describe this on <span class="math-container">$S_3$</span>:</p>
<ul>
<li>the identity switches with the involution 321.</li>
<li>the transpositions 213 and 132 switch.</li>
<li>the 3-cycles 312 and 231 switch.</li>
</ul>
<p>In general, this operation preserves being order <span class="math-container">$\leq 2$</span> (since this is equivalent to the P- and Q-symbols being the same).</p>
| darij grinberg | 2,530 | <p>Edit: This doesn't answer the question; see comments.</p>
<p>Knuth, TAoCP3, p. 76, exercise 5 in section 5.1.4:</p>
<p>Let $P$ be the tableau corresponding to the permutation $a_1a_2...a_n$; use exercise 4 to prove that $P^T$ is the tableau corresponding to $a_n...a_2a_1$.</p>
|
14,527 | <p>The <a href="https://en.wikipedia.org/wiki/Robinson%E2%80%93Schensted%E2%80%93Knuth_algorithm" rel="nofollow noreferrer">Robinson-Schensted correspondence</a> is a bijection between elements of the symmetric group and ordered pairs of standard tableaux of the same shape.</p>
<p>Some simple operations on tableaux correspond to simple operations on the group: switching the tableaux corresponds to inverse on the group.</p>
<blockquote>
<p>What about taking the transpose of the tableaux? Does that correspond to something easily described on permutations?</p>
</blockquote>
<p>In order to rule out easy guesses, let me describe this on <span class="math-container">$S_3$</span>:</p>
<ul>
<li>the identity switches with the involution 321.</li>
<li>the transpositions 213 and 132 switch.</li>
<li>the 3-cycles 312 and 231 switch.</li>
</ul>
<p>In general, this operation preserves being order <span class="math-container">$\leq 2$</span> (since this is equivalent to the P- and Q-symbols being the same).</p>
| Richard Stanley | 2,807 | <p>It doesn't seem natural to the theory of RSK to transpose both tableaux. When we apply
RSK to <span class="math-container">$a_n\cdots a_1$</span> then we get the pair <span class="math-container">$(P^t,\mathrm{evac}(Q)^t)$</span>, where evac is
Schutzenberger's evacuation operation. See for instance <em>Enumerative Combinatorics</em>, vol. 2, pp. 425--429. For more on evacuation, see <a href="https://doi.org/10.37236/75" rel="nofollow noreferrer">https://doi.org/10.37236/75</a>.</p>
|
183,058 | <p>I'm having difficulty producing a proper animated gif (for inclusion in a PowerPoint presentation). The code I have is this:</p>
<pre><code> HSBhsl[h_, s_, l_] :=
Module[{H, S, B}, H = h; B = l + s (1 - Abs[2 l - 1])/2;
S = (2 B - l)/B; {H, S, B}]
TestCube[\[Theta]_] :=
Graphics3D[Rotate[{EdgeForm[], Hue[HSBhsl[32/250, 228/250, 105/256]],
Cuboid[{0, 0, 0}, {1, 1, 1}]}, -\[Theta] Degree, {0, 0, 1}, {1, 1,
1}], Lighting -> {{"Ambient",
RGBColor[0.7, 0.7, 0.7]}, {"Directional",
RGBColor[0.35, 0.35, 0.35], ImageScaled[{0, 5, 0}]}},
Boxed -> False, Background -> None,
PlotRange -> 4 {{-1, 1}, {-1, 1}, {-1, 1}}, ViewAngle -> Pi/10000]
</code></pre>
<p>The animation is then produced like so:</p>
<pre><code>list = Table[
Show[TestCube[t], ViewPoint -> {1000, 1000, 1000}], {t, 0, 360, 1}];
SetDirectory@NotebookDirectory[];
Export["CubesAnimation.gif", list, "TransparentColor" -> White]
</code></pre>
<p>Unfortunately, this produces an amimation that seems to stack images on top of each other, thus giving the appearance of a "trace" left by the previous frame. I tried adding <code>"TransitionEffect" -> Background</code>, but this had no effect.</p>
<p>How do I get a proper GIF animation, with a transparent background, of the rotating cube?</p>
| rafalc | 51,110 | <p>I have investigated the problem and it seems that this is indeed a limitation of <code>Export</code>. We will add support for <code>"TransitionEffect"</code> option in <code>Export</code> in the next release of Mathematica.</p>
<p>In the meantime, as a workaround, you can use a low-level function from GIFTools paclet <code>GIFTools`Private`$WriteFrames</code>. The function is not documented because it is not supposed to be used directly, it serves as a back-end to <code>Export</code>.</p>
<p>But here is how you can use it in your case:</p>
<pre><code>list = Table[Show[TestCube[t], ViewPoint -> {1000, 1000, 1000}], {t, 0, 360, 1}];
SetDirectory@NotebookDirectory[];
(* Load and initialize GIFTools paclet *)
Needs["GIFTools`"]
GIFTools`Private`InitGIFTools[]
GIFTools`Private`$WriteFrames[
Image3D[list], (* All frames stuffed into a single Image3D *)
"CubesAnimation.gif", (* Output file name *)
"", (* Comment, none in this case *)
{0.1}, (* Display duration for each frame *)
"None", (* Dithering method, none in this case *)
1, (* How many times to play the animation *)
True, (* Interlacing *)
{255, 255, 255}, (* Background color *)
{255, 255, 255}, (* Transparent color - white *)
{2} (* Disposal mode - Restore to background color*)
]
</code></pre>
<p>As an alternative you can consider exporting to Animated PNG, which supports alpha channel.</p>
|
4,322,897 | <p>Let <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> be a sequence in <span class="math-container">$\Bbb{R}^n$</span> and <span class="math-container">$z\in\Bbb{R}^n$</span> be given. If <span class="math-container">$\{\|z-y_n\|\}_{n=1}^\infty$</span> is a convergent sequence, what can we say about <span class="math-container">$\{y_n\}_{n=1}^\infty$</span>? More precisely, can we conclude that <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> is bounded?</p>
<p>For some reason, I'd like to extract a convergent subsequence from <span class="math-container">$\{y_n\}_{n=1}^\infty$</span>, and this can be done by showing that <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> is bounded. Here is my attempt. Since <span class="math-container">$\{\|z-y_n\|\}_{n=1}^\infty$</span> converges, it must be bounded. Then <span class="math-container">$\exists r>0$</span> s.t. <span class="math-container">$\forall n\in\Bbb N$</span>, <span class="math-container">$\left|\|z-y_n\|\right|<r$</span>, but this amounts to saying that each term of <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> falls into an open ball centered at <span class="math-container">$z$</span>. Thus, <span class="math-container">$\{y_n\}_{n=1}^\infty$</span> must be a bounded sequence. Is my attempt correct, please? Thank you.</p>
| LGu | 987,256 | <p>First let the sequence WLOG be unbounded from above, i.e. for each <span class="math-container">$z\in R$</span>, <span class="math-container">$\exists k$</span> such that <span class="math-container">$y_k>z$</span>. Now let <span class="math-container">$\epsilon>0$</span> be given, if <span class="math-container">$|z-y_n|$</span> converges then there is some <span class="math-container">$m$</span> such that for all <span class="math-container">$q>m$</span>, <span class="math-container">$|z-y_q|\leq \epsilon$</span>, which implies that <span class="math-container">$y_q\leq z+\epsilon$</span>.<br />
But this contradicts with the unboundedness of the sequence <span class="math-container">$y_n$</span>, since the finite set <span class="math-container">$\{y_o\},o=1,…,m$</span> is definitely bounded, and for each <span class="math-container">$j>m$</span>, <span class="math-container">$y_j\leq z+\epsilon$</span>. Let <span class="math-container">$b=max(sup\{y_o\},z+\epsilon)$</span>, we can’t find <span class="math-container">$k$</span> such that <span class="math-container">$y_k>b$</span>.</p>
|
2,489,429 | <p>$\int \frac{x}{x+1}\,dx$ so I set substitution to $x+1=t$ then I differentiate to get $dx=dt$ and then proceed to get $\int 1\,dt - \int \frac{1\,dt}{t}$ to get $t-\ln(|t|)+C$ but when I replace back the substitution I get $x+1$ in the final answer instead of just the $x$ , so my question is why was it a mistake to integrate $\int 1\,dt$ instead to switch back to x and then integrate $\int 1\,dx$</p>
| Cornman | 439,383 | <p>Alternative:</p>
<p>$\int \frac{x}{x+1}\, dx=\int \frac{x+1-1}{x+1}\, dx=\int 1-\frac{1}{x+1}\, dx= x-\ln|x+1|+C$</p>
<p>Without substitution.</p>
|
153,999 | <p>How to solve: $|\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1$.</p>
<p>I would like to know how to solve an absolute value equation when there is a square root sign inside.</p>
| Steven Alexis Gregory | 75,410 | <p>Theorem: Let $X,P,Q \in \mathbb R^n$. Then $X \in \overline{PQ} \iff d(P,X) + d(X,Q) = d(P,Q)$</p>
<p>On the set of real numbers, the distance between numbers $x$ and $y$ is $|x-y|$. So we have this.</p>
<p>Corollary: Let $x,p,q \in \mathbb R$ with $p < q$. Then $p \le x \le q \iff |x-p| + |q-x| = q-p$</p>
<p>In this case, </p>
<p>\begin{align}
|\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1
&\iff 2 \le \sqrt{x-1} \le 3 \\
&\iff 4 \le x-1 \le 9 \\
&\iff 5 \le x \le 10
\end{align}</p>
|
3,114,663 | <p>Let there be <span class="math-container">$m$</span> indistinguishable balls, <span class="math-container">$k$</span> bins, <span class="math-container">$C$</span> capacity.
Let <span class="math-container">$X_j$</span> denote the total balls in bin <span class="math-container">$j$</span>.
I've seen ways to calculate the total number of combinations, but I'm not sure how to go about calculating the mean and variance of <span class="math-container">$X_j$</span>.
It is understood that the ball is always thrown into one of the empty bins with uniform probability. edit: nonfull bins not empty bins sorry.</p>
| parsiad | 64,601 | <p>Here's some a Monte Carlo simulation in Python for the problem. Disclaimer: it's not efficient.</p>
<pre><code>import numpy as np
import sys
def simulate(num_balls, num_bins, capacity, num_sims):
assert(capacity * num_bins >= num_balls)
results = np.empty((num_sims,))
for i in xrange(num_sims):
unfilled_bins = set(range(num_bins))
bin_counts = np.zeros((num_bins,))
for _ in xrange(num_balls):
chosen_bin = np.random.choice(list(unfilled_bins))
bin_counts[chosen_bin] += 1
if bin_counts[chosen_bin] >= capacity:
unfilled_bins.remove(chosen_bin)
if chosen_bin == 0:
break # Stop the simulation prematurely.
results[i] = bin_counts[0]
return results
if __name__ == '__main__':
try:
num_balls, num_bins, capacity, num_sims = map(int, sys.argv[1:])
except:
sys.stderr.write(('usage: {} NUM_BALLS NUM_BINS CAPACITY '
'NUM_SIMS\n').format(sys.argv[0]))
sys.exit(1)
total_capacity = capacity * num_bins
if total_capacity < num_balls:
sys.stderr.write(('error: parameters must satisfy'
'CAPACITY * NUM_BINS >= NUM_BALLS`.\n'))
sys.exit(2)
results = simulate(num_balls, num_bins, capacity, num_sims)
mean = results.mean()
var = results.var()
sys.stdout.write('mean = {}\tvar = {}\n'.format(mean, var))
<span class="math-container">```</span>
</code></pre>
|
1,928,892 | <p>I'm solving a graduate entrance examination problem.
We are required to establish the inequality using the following result:</p>
<p>for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x - y)^2 > 0$.</p>
<p>But when it comes to an inequality combining $x, y, z$, I got stuck as I've tried to develop the expression into one single fraction and obtain something irreducible.</p>
<p>Any hints ? My intuition tells me that for $x,y,z >0$, any fraction of the form $\frac{x}{y+z}$ is greater than 1/2. As there are three fractions of this kind with mute variables playing symmetrical roles, we get: $1/2 + 1/2 + 1/2 = 3/2$.</p>
<p>I just don't figure out how to play with the result (1).</p>
| Zuo | 399,908 | <p>Proof:</p>
<p>If $x,y,z>0$,then:$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$$$\geq3(xy+yz+zx)$$</p>
<p>So:$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}$$$$=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{yz+zx}$$$$\geq\frac{(x+y+z)^2}{2(xy+yz+zx)}\geq\frac3{2}$$</p>
|
2,736,596 | <p>I want to proof this equation below.
$$
f^{-1}(\bigcup B_{j})=\bigcup(f^{-1}B_{j})
$$
I know it's true but I don't have idea how to proof and explain.
Please help.</p>
| drhab | 75,923 | <p>Essential is: $$x\in f^{-1}(A)\iff f(x)\in A$$</p>
<p>So the following statements are equivalent:</p>
<ul>
<li>$x\in f^{-1}(\bigcup_{j\in J}B_j)$</li>
<li>$f(x)\in\bigcup_{j\in J}B_j$</li>
<li>$f(x)\in B_{j_0}$ for some $j_0\in J$</li>
<li>$x\in f^{-1}(B_{j_0})$ for some $j_0\in J$</li>
<li>$x\in\bigcup_{j\in J}f^{-1}(B_j)$</li>
</ul>
|
2,487,044 | <p>Why are integral elements defined in terms of monic polynomials? Why do we wish to split the cases between non monic polynomials and monic polynomials? I.e, algebraic elements and integral elements. Motivation? </p>
| reuns | 276,986 | <p>Try with $\sqrt{2}$ then the ring $\mathbb{Z}[\sqrt{2}]$ is just $\{ a+b\sqrt{2}, (a,b) \in \mathbb{Z}\}$. With $\frac{\sqrt{2}}{2}$ then the ring $\mathbb{Z}[\frac{\sqrt{2}}{2}]$ is $\{\frac{a+b\sqrt{2}}{2^k}, (a,b,k) \in \mathbb{Z}\}$. The difference is that $\frac{\sqrt{2}}{2}$ is not the root of any monic polynomial in $\mathbb{Z}[x]$.</p>
<p>In other words $\mathbb{Z}[\alpha]$ is a finitely generated $\mathbb{Z}$-module iff $\alpha$ is the root of a monic polynomial.</p>
|
2,990,400 | <p>I need to find all the continuous functions <span class="math-container">$f$</span> such that
<span class="math-container">$$
[f(t)]^2=\int_0^t f(s) ds
$$</span>
<strong>Attempt:</strong></p>
<p>Since
<span class="math-container">$$
f(t)\leq [f(t)]^2+c, \, \, \forall c \geq \frac14$$</span>
we have</p>
<p><span class="math-container">$$
f(t)\leq c+ \int_0^t f(s)ds $$</span>
Now, Gronwall's Lemma gives
<span class="math-container">$$
f(t) \leq c+ \int_0^t c \,e^{t-\tau}d\tau \iff$$</span>
<span class="math-container">$$ f(t) \leq c \, e^t, \, \,\forall c\geq \frac14$$</span></p>
<p>Given that these steps were correct, is this inequality the final answer?</p>
| TurlocTheRed | 397,318 | <p>You can use the Lemma use to prove the Fundamental Theorem of Calculus:</p>
<p><span class="math-container">$$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$</span></p>
<p>Here we have <span class="math-container">$[f(t)]^2=\int_0^tf(s)ds.$</span></p>
<p>Take the <span class="math-container">$t$</span> derivative of both sides. </p>
<p>So <span class="math-container">$2f(t)f'(t)-f(t)=0=f(t)[2f'(t)-1].$</span></p>
<p>So either <span class="math-container">$f(t)=0$</span>, or <span class="math-container">$2f'(t)-1=0\implies f(t)=t/2+c$</span>.</p>
|
2,238,165 | <p>Can anyone help with this limit?</p>
<p>\begin{equation*}
\lim_{x \rightarrow 4}
\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16(x-4)^2}
\end{equation*}</p>
<p>I've tried a variable change of \begin{equation*} y=\sqrt{x} \end{equation*} but this didn't help.</p>
| Especially Lime | 341,019 | <p>You can almost calculate it without any cases, but there is a slight extra complication from the fact that the first digit can't be $0$.</p>
<p>If you just wanted to know how many strings of $N$ digits with exactly two $5$s and three $7$s, then you can first choose two places from $N$ for the $5$s, then three places (from the remaining $N-2$) for the $7$s. Then you have $N-5$ places to fill, and each one can be any of the other $8$ digits. So that gives $\binom N2\binom{N-2}38^{N-5}$
ways.</p>
<p>Unfortunately this includes things like $05787735$, which you don't want. So you need to subtract off the number of strings which start with a $0$, and have two $5$s and three $7$s. You should be able to use the same method to calculate how many strings of this form there are.</p>
|
3,245,854 | <p>Hi I am trying to solve an integral problem that involves trig substitution. First I tried completing the square, which gave me <span class="math-container">$1/\sqrt{(x+3)^2+2^2}$</span>. I know I am supposed to use <span class="math-container">$x = \arctan(\theta)$</span>. Does that mean it should be: </p>
<p><span class="math-container">$x + 3 = \arctan(\theta)$</span></p>
<p><span class="math-container">$x = \arctan(\theta) - 3$</span></p>
<p>and then Integrate from there? I am not sure if this is a good way of thinking about this problem. I would appreciate any help!
<a href="https://i.stack.imgur.com/N2IV6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N2IV6.png" alt="enter image description here"></a></p>
| John Wayland Bales | 246,513 | <p><span class="math-container">$$\int \frac{1}{\sqrt{x^2+6x+13}}\thinspace dx=\int\frac{1}{\sqrt{(x+3)^2+2^2}}\thinspace dx$$</span>
<a href="https://i.stack.imgur.com/lv0K4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lv0K4.png" alt="Trig substitution"></a></p>
<p>So <span class="math-container">$x+3=2\tan\theta$</span>, and <span class="math-container">$dx=2\sec^2\theta\thinspace d\theta$</span>
<span class="math-container">\begin{eqnarray} \int\frac{1}{\sqrt{(x+3)^2+2^2}}\thinspace dx&=&=\frac{1}{2}\int\frac{2}{\sqrt{(x+3)^2+2^2}}\thinspace dx\\\
&=&\frac{1}{2}\int\cos(\theta)\sec^2(\theta)\thinspace d\theta\\\
&=&\frac{1}{2}\int\sec\theta\thinspace d\theta
\end{eqnarray}</span></p>
<p>After completing the integration, substitute</p>
<p><span class="math-container">$$\theta=\arctan\left(\frac{x+3}{2}\right)$$</span>.</p>
|
131,583 | <p>My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any <strong>algorithm</strong>) that we can use to rewrite it using language of differential forms, tensor, exterior calculus, Hodge star and other operators which are coordinate independent? An example, the <strong>Grad f</strong> can be rewritten as a geometric form: (df)#, where # is a sharp operator turning a one-form into a vector. I am currently facing this problem to turn a partial differential equation into its coordinate-independent form, which involves forms, tensors, exterior calculus and other operators. </p>
<p>Thank you for anyone who help me about this problem!</p>
| Victor Lehenkyi | 39,966 | <p>Basic paper is
Journal of Mathematical Physics , Vol. 12. New York: American Institute of Physics. (1971): 653-666 (so called "Harrison-Estabrook procedure.")
This procedure is implemented in package "Liesymm" for Maple.</p>
|
551,471 | <p>In a proof I am reading, the author considers the situation where $G$ is a reductive algebraic group (variety) over the complex numbers $\mathbb C$ and $N\trianglelefteq G$ is a closed, normal subgroup of $G$ which is also finite. The author concludes that <em>"because $G$ is reductive"</em>, $N$ must be contained in the center of $G$. I can't seem to make that same deduction, though. Is this some well-known result? Can you give a proof?</p>
<p>Btw, if $G$ is connected then this does not seem to have anything to do with reductivity at all. Let $u\in N$, then $G.u\subseteq N$ is finite. Since $0=\dim(G.u)=\dim(G)-\dim(G_u)$, the stabilizer of $u$ is a closed subgroup of $G$ which has maximal dimension, therefore $G=G_u$. However, the author explicitly states that the reductivity of $G$ is the reason for his deduction, and I am curious what that would be.</p>
| user466484 | 466,484 | <p>Actually, any finite normal subgroup H of a connected algebraic group G is central. (Consider the action GxH->H by conjugation. The orbits are connected.) </p>
|
2,077,053 | <p>Please check my proof</p>
<p>If $H\triangleleft G$ then $gHg^{-1}=H $ for every $g\in G $</p>
<p>If $K\triangleleft G$ then $gKg^{-1}=K$ for every $g\in G$</p>
<p>Then $H\cap K = gHg^{-1}gKg^{-1}=gHKg^{-1}$</p>
<p>Then $H\cap K\triangleleft G$</p>
| Zelos Malum | 197,853 | <p>You proof is flawed as
$$H\cap K \subseteq H\subseteq HK$$
where equality does not need to hold so you cannot say that.</p>
<p>A better method is this to show that for $x\in H\cap K$ we have $gxg^{-1}\in H\cap K$ for all $g$.</p>
<p>This is trivial as $x\in H$ gives that $gxg^{-1}\in H$ and $x\in K$ gives that $gxg^{-1}\in K$ which means that $gxg^{-1}\in H\cap K$.</p>
|
2,254,694 | <p>I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:</p>
<p>Attempt 1:</p>
<p>Write out the terms of $b_n$:</p>
<p>$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$</p>
<p>Adding up the terms you get:</p>
<p>$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$</p>
<p>But a dead end here.</p>
<p>Attempt 2:</p>
<p>For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$</p>
<p>Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. </p>
<p>For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. </p>
| Michał Miśkiewicz | 350,803 | <p>Let $c_n = 2^n a_n$. The assumption reads
$$ \frac{c_{n+1}-c_n}{2^{n+1}-2^n} \to 0. $$
The sequence $2^n$ tends monotonically to $+\infty$, hence by the <a href="https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem" rel="nofollow noreferrer">Stolz–Cesàro theorem</a>
$$ a_n = \frac{c_n}{2^n} \to 0. $$</p>
|
1,517,245 | <p>I would appreciate if somebody could help me with the following problem</p>
<p>Q: Arrangements of $1,2,..,81$ in an $9 \times 9$ matrix such that each row and each column is increasing. How many possible $X$ ?</p>
<p><a href="https://i.stack.imgur.com/Te1TB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Te1TB.png" alt="enter image description here"></a></p>
| Martin Brandenburg | 1,650 | <p>They are sometimes also called <strong>pointed monoids</strong>. See <a href="https://arxiv.org/abs/1909.00297v1" rel="nofollow noreferrer">arXiv:1909.00297</a> for instance.</p>
|
2,008,243 | <p>How to find the exact value of the following limit?</p>
<p>$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$</p>
| Nicolas FRANCOIS | 288,125 | <p>$$\cos x^2=1-\frac{x^4}{2}+o(x^4)$$
so
$$e^{x\cos x^2} = e^{x-\frac{x^5}{2}+o(x^5)} = e^xe^{-\frac{x^5}{2}+o(x^5)}$$
therefore
$$\frac{e^{x\cos x^2}-e^x}{x^5} = e^x\frac{e^{-\frac{x^5}{2}+o(x^5)}-1}{x^5} \sim e^x\frac{-\frac{x^5}{2}}{x^5} \xrightarrow[x\to0]{}-\frac{1}{2}$$
using $e^u-1\sim u$ when $u\to0$.</p>
|
5,581 | <p>(Wade's rewording)</p>
<p>There is a phrase that contains $4$ words and $22$ total letters (not counting the three spaces between the words). One letter appears four times, no letter appears three times, three letters appear twice, and $12$ letters appear once each. What is this phrase?</p>
<blockquote>
<p>@@$$##&&&%%%%????????????</p>
</blockquote>
<p>There are $4$ words and the & represents a whitespace.<br>
@,#,\$,% represent one letter each ($4$ total letters).<br>
Each ? represents one of the other $22$ letter of the alphabet ($12$ total letters).<br>
$16$ different letters used.<br>
ciphertext also needs to be rearranged - it is 'scrambled.' </p>
<p>How do I approach this?<br>
Not looking for the answer really.<br>
I just need a tool to unscramble the phrase and decrypt the ciphertext simultaneously.<br>
Decrypto only does the later. Thank you. </p>
<p>@Noldorin: This is not homework. From elsewhere. </p>
<p>@Kenny: How are encryption/decryption methodologies not math? This is an applied math problem, whereas your preferences tend to be theoretical/ pure math?</p>
| David E Speyer | 448 | <p>Captain Kirk was wondering why his journal was not more excited to hear about his exploits. He said</p>
<p>LOG, YOU SEEM FAINTHEARTED!</p>
<hr>
<p>More seriously, I dumped the letters EEEETTAAOOINSHRDLUGFMY into a multiword anagrammer, and it immediately found more than 24000 answers, of which this was one of the more amusing ones. What I deduce from this experience is that, without the knowledge that you are looking for a common English phrase, this puzzle is dramatically under constrained. </p>
<p>I recommend you look for a file of common English phrases which you can test for this property. </p>
<p>This isn't math, but there is a bit of a mathematical lesson here -- it is worth doing a quick test to see whether you have enough data to answer a question before working too hard on it.</p>
|
578,086 | <p>Hey I've been having problem finding the limit of the sequence below</p>
<p>$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}$</p>
<p>Thanks!</p>
| gangrene | 103,440 | <p>The solution I know involves the <em>p-norms</em> and the <em>Gamma function</em>: in fact let $f: [0, \infty[ \to \mathbb{R}$ be defined by $f(x)=xe^{-x}$. Since $f$ is positive, continuous, and $0=f(0)=f(\infty)$, we have $\|f\|_\infty = \max f([0,\infty])$; now $f'(x)=(1-x)e^{-x}$ is $0$ for $x=1$, so that $\|f\|_\infty = f(1)=1/e$. Moreover $$\int_0^\infty |f(x)|^p \ dx=\int_0^\infty x^p e^{-px} dx = \int_0^\infty \frac{t^p}{p^p} e^{-t} \frac{dt}{p} = \frac{\Gamma(p+1)}{p^{p+1}}$$ then $$\|f\|_p = \frac{(\Gamma(p+1))^{1/p}}{p p^{1/p}}$$ since $\lim_{p \to \infty} \|f\|_p = \|f\|_{\infty} = 1/e$ and $\lim_{p \to \infty} p^{1/p}=1$, we get $$\lim_{p \to \infty} \frac{(\Gamma(p+1))^{1/p}}{p p^{1/p}} = \frac{1}{e} \Longrightarrow \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=\frac{1}{e} $$</p>
<p>I hope that this will help.</p>
|
1,279,947 | <p>In physics, 1m is defined 'the length which light travels in certain time'.
And i've thought 2m has twice many elements than 1m.
But today i've realized intervals (0,1) and (0,2) are equipotent.
I was shocked because i've thought length and cardinality are similar concepts.
Cardinality is clearly defined in set theory. But what is length? </p>
| 5xum | 112,884 | <p>The best "approximation" of length in mathematics is "measure theory". This is an extensive field of mathematics that deals with "sizes" (called "measures") of sets. </p>
<p>You can define a measure on $\mathbb R$ on which the measure of an interval $(a,b)$ is equal $b-a$. This is called the Lebesgue measure, and it fits what you understand as "twice as big", since the measure of $[0,2]$ is two times the measure of $[0,1]$. The Lebesgue measure can also be defined on $\mathbb R^n$, and there, the size of the "rectangle" $$[a_1,b_1]\times[a_2,b_2]\times\cdots\times [a_n, b_n]$$</p>
<p>is equal to $$(b_1-a_1)\cdot (b_2-a_2)\cdots (b_n-a_n).$$</p>
<p>For initial further reading, look up the <a href="http://en.wikipedia.org/wiki/Measure_%28mathematics%29" rel="nofollow">wikipedia page on measure theory</a></p>
|
28,586 | <p>Imagine I have two lists:</p>
<pre><code>List1 = {0, 1, 0.6, 0.5, 1.2, 0.4};
List2 = {"a", "b", "c", "d", "e", "f"};
</code></pre>
<p>How can I use <code>Pick</code> to return a third list consisting of items in List2 that have a array position corresponding to an array position with a value $\leq N$ in the first list? Here, for example, for $N = 0.5$ we should have:</p>
<pre><code>List3 = {"a", "d", "f"}
</code></pre>
| chyanog | 2,090 | <p>You can also use <code>UnitStep</code>, it has the Listable attribute, listable operations will be faster than other alternatives.</p>
<pre><code>List1 = {0, 1, 0.6, 0.5, 1.2, 0.4};
List2 = {"a", "b", "c", "d", "e", "f"};
Pick[List2, UnitStep[0.5 - List1], 1]
(*{"a", "d", "f"}*)
</code></pre>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| robjohn | 13,854 | <p>Compute the generating function of the harmonic numbers:
$$
\begin{align}
\sum_{n=1}^\infty H_nx^n
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac{x^n}{k}\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{x^n}{k}\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{x^{n+k}}{k}\\
&=-\frac{\log(1-x)}{1-x}\tag{1}
\end{align}
$$
Integrating $(1)$ yields
$$
\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n=\frac12\log(1-x)^2\tag{2}
$$
Therefore,
$$
\sum_{n=1}^\infty\frac{H_{n-1}}{n}e^{\pm2\pi inx}=\frac12\log(1-e^{\pm2\pi ix})^2\tag{3}
$$
Multiplying and integrating gives
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}
&=\frac14\int_0^1\log(1-e^{2\pi ix})^2\log(1-e^{-2\pi ix})^2\,\mathrm{d}x\tag{4a}\\
&=\frac1{8\pi i}\oint\log(1-z)^2\log(1-1/z)^2\frac{\mathrm{d}z}{z}\tag{4b}\\
&=\frac1{8\pi i}\int_0^1\log(1-z)^2\left[-\pi i+\log(1-z)-\log(z)\right]^2\frac{\mathrm{d}z}{z}\\
&-\frac1{8\pi i}\int_0^1\log(1-z)^2\left[\pi i+\log(1-z)-\log(z)\right]^2\frac{\mathrm{d}z}{z}\tag{4c}\\
&=-\frac12\int_0^1\log(1-z)^2\left[\log(1-z)-\log(z)\right]\frac{\mathrm{d}z}{z}\tag{4d}
\end{align}
$$
Explanation<br>
$\mathrm{(4a)}$: multiply the conjugates of $(3)$ and integrate<br>
$\mathrm{(4b)}$: convert to contour integral with $z=e^{2\pi ix}$<br>
$\mathrm{(4c)}$: deflate the contour to lines above and below $[0,1]$<br>
$\mathrm{(4d)}$: algebra</p>
<p>Contour $\color{#00A000}{\text{before}}$ and $\color{#C00000}{\text{after}}$ $\mathrm{(4c)}$:</p>
<p>$\hspace{4cm}$<img src="https://i.stack.imgur.com/hve8l.png" alt="deflate the contour to lines above and below $[0,1]$"></p>
<p>Using $\log(1-z)=-u$, we get
$$
\begin{align}
\int_0^1\log(1-z)^3\frac{\mathrm{d}z}{z}
&=-\int_0^\infty u^3\frac{\mathrm{d}u}{e^u-1}\\
&=-\Gamma(4)\zeta(4)\\
&=-\frac{\pi^4}{15}\tag{5}
\end{align}
$$
Using $\log(z)=-u$ and , we get
$$
\begin{align}
\int_0^1\log(1-z)^2\log(z)\frac{\mathrm{d}z}{z}
&=-\int_0^\infty\log(1-e^{-u})^2u\,\mathrm{d}u\tag{6a}\\
&=-2\sum_{n=1}^\infty\int_0^\infty\frac{H_{n-1}}{n}e^{-nu}u\,\mathrm{d}u\tag{6b}\\
&=-2\sum_{n=1}^\infty\frac{H_{n-1}}{n^3}\tag{6c}\\
&=\zeta(2)^2-3\zeta(4)\tag{6d}\\
&=-\frac{\pi^4}{180}\tag{6e}
\end{align}
$$
Explanation<br>
$\mathrm{(6a)}$: substitute $z=e^{-u}$<br>
$\mathrm{(6b)}$: apply $(2)$<br>
$\mathrm{(6c)}$: integrate<br>
$\mathrm{(6d)}$: use <a href="https://math.stackexchange.com/a/469785">this answer</a><br>
$\mathrm{(6e)}$: evaluate</p>
<p>Combining $(4)$, $(5)$, and $(6)$ yields
$$
\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}=\frac{11\pi^4}{360}\tag{7}
$$
Noting that
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}
&=\sum_{n=1}^\infty\frac{\left(H_n-\frac1n\right)^2}{n^2}\\
&=\sum_{n=1}^\infty\left(\frac{H_n^2}{n^2}-2\frac{H_n}{n^3}+\frac1{n^4}\right)\tag{8}
\end{align}
$$
we get, again using <a href="https://math.stackexchange.com/a/469785">this answer</a>, that
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_n^2}{n^2}
&=\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}+2\sum_{n=1}^\infty\frac{H_n}{n^3}-\zeta(4)\\
&=\frac{11\pi^4}{360}+5\zeta(4)-\zeta(2)^2-\zeta(4)\\
&=\frac{17\pi^4}{360}\tag{9}
\end{align}
$$</p>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| r9m | 129,017 | <p>Starting with $\displaystyle \dfrac{H_n}{n} = \sum_{k=1}^{\infty} \dfrac{1}{k(k+n)}$ we have,
\begin{align*}\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} &= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2\\&= \sum_{n=1}^{\infty}\sum_{k,j=1}^{\infty} \dfrac{1}{jk(j+n)(k+n)} \\&= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty} \dfrac{1}{k^2(n+k)^2}+ 2\sum_{1 \le k<j} \dfrac{1}{jk(j+n)(k+n)}\right) \\&= \sum_{1 \le k < j} \dfrac{1}{k^2j^2} + 2\sum_{n=1}^{\infty}\sum_{k,m=1}^{\infty} \dfrac{1}{k(k+m)(k+m+n)(k+n)}\\&= \dfrac{1}{2}\left(\left(\sum_{k=1}^{\infty}\dfrac{1}{k^2}\right)^2 – \sum_{k=1}^{\infty}\dfrac{1}{k^4}\right) + 2\sum_{k,m,n=1}^{\infty}\dfrac{(k+m)(k+n) – k(k+m+n)}{kmn(k+m)(k+m+n)(k+n)} \\&= \dfrac{1}{2}\left(\zeta^2(2) – \zeta(4)\right) + 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} – 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{mn(m+k)(n+k)}\end{align*}</p>
<p>Therefore, $\displaystyle 3\sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2 = \dfrac{1}{2}\left(\zeta^2(2) - \zeta(4)\right) + 2\sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)}$</p>
<p>Using, $\displaystyle \sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} = 6\zeta(4)$ we conclude, </p>
<blockquote>
<p>$$\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} = \dfrac{1}{6}\zeta^2(2) + \dfrac{23}{6}\zeta(4)$$</p>
</blockquote>
<p>To see the last result, \begin{align*} \sum_{k,m,n=1}^{\infty} \dfrac{1}{kmn(k+m+n)} &= \sum_{k,m,n=1}^{\infty} \int_0^1 \dfrac{x^{k+m+n}}{kmn}\,\dfrac{dx}{x} \\&= \int_0^1 \log^3(1-x)\,\dfrac{dx}{x} \\&= -\int_0^1 \dfrac{\log^3 x}{1-x}\,dx \\&= -\sum_{n=0}^{\infty} \int_0^1 x^n\log^3 x\,dx \\&= 6\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^4} = 6\zeta(4)\end{align*}</p>
|
3,081,552 | <p>Is this <code>∅ ⊈ { ∅, 1, 2 }</code> true or false ? </p>
<p>Also, I am confuse since this <code>{ ∅, 1, 2 }</code> has already contain a <code>∅</code>, does it still contain another <code>∅</code> meaning like : <code>{ ∅, ∅, 1, 2 }</code> ? </p>
<p>Is <code>∅ ∈ { ∅, 1, 2 }</code> true ? & <code>{∅} ∈ { ∅, 1, 2 }</code> false ?</p>
| Anik Bhowmick | 354,636 | <p>No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.</p>
|
2,390,804 | <p>How does one compute the following integral?
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$</p>
<hr>
<p>I have tried extending $x$ to the complex plane then evaluating the following contour integral
$$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$
with the contour $C$ running along the whole real axis and then upper semicircle. I obtain
$$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$
but not the original integral.</p>
| Random Variable | 16,033 | <p>Using the Laplace transform of the sine function, we get <span class="math-container">$$ \begin{align} \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx &= \int_{0}^{\infty} \sqrt{x} \sin(x) \int_{0}^{\infty} \sin(t) e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \int_{0}^{\infty} \sin(t) \int_{0}^{\infty} \sqrt{x} \sin(x) e^{-tx} \, \mathrm dx \, \mathrm dx \\ &= -\Im\int_{0}^{\infty} \sin(t) \int_{0}^{\infty} \sqrt{x} e^{-(t+i)x} \, dx \, \mathrm dt \\ &= -\frac{\sqrt{\pi}}{2} \, \Im \int_{0}^{\infty} \frac{\sin(t)}{(t+i)^{3/2}} \, \mathrm dt \tag{1}\\ &= - \sqrt{\pi} \, \Im \int_{0}^{\infty} \frac{\cos (t)}{\sqrt{t+i}} \, \mathrm dt \tag{2}\\ &=- \frac{\sqrt{\pi}}{2} \, \Im \left(\int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt + \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt \right). \end{align} $$</span></p>
<hr />
<p><span class="math-container">$(1)$</span> <span class="math-container">$\mathcal{L} \{\sqrt{x} \} (s) = \frac{\sqrt{\pi}}{2 s^{3/2}}$</span> if <span class="math-container">$\Re(s) >0$</span></p>
<p><span class="math-container">$(2)$</span> Integrate by parts.</p>
<hr />
<p>Next, using the principal branch of the square root, let's integrate the function <span class="math-container">$$f(z) = \frac{e^{iz}}{\sqrt{z+i}}, $$</span> around an infinitely large closed quarter-circle in the first quadrant of the complex plane.</p>
<p>Applying Jordan's lemma, we get</p>
<p><span class="math-container">$$ \begin{align} \int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt&=e^{i \pi/4} \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+1}} \, \mathrm dt \\ &= 2 e^{i \pi/4} \, e \int_{1}^{\infty} e^{-u^{2}} \, \mathrm du \tag{3} \\ &= 2 e^{i \pi/4} e \, \frac{\sqrt{\pi}}{2} \operatorname{erfc} (1) \\ &= e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc} (1). \end{align}$$</span></p>
<p>And by integrating the function <span class="math-container">$$g(z) = \frac{e^{-iz}}{\sqrt{z+i}} $$</span> around an infinitely large closed quarter-circle in the fourth quadrant of the complex plane (indented at <span class="math-container">$z=-i$</span>), we get <span class="math-container">$$ \begin{align} \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt &= \int_{0}^{1} \frac{e^{-t}}{\sqrt{(1-t)e^{ i \pi /2}}} \, (-i \, \mathrm dt) + \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{(t-1)e^{-i \pi /2} }} \, (-i \mathrm \, dt) \\ &= -e^{i \pi/4}\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm dt + e^{ - i \pi /4} \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{t-1}} \, \mathrm dt \\ &= -\frac{2 e^{ i \pi /4}}{e} \int_{0}^{1} e^{v^{2}} \, \mathrm dv + \frac{2e^{ - i \pi/4}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \tag{4} \\ &= -\frac{2e^{ i \pi /4}}{e} \frac{\sqrt{\pi}}{2} \operatorname{erfi}(1) + \frac{2 e^{ - i \pi /4}}{e}\frac{\sqrt{\pi}}{2} \\ &= \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right). \end{align}$$</span></p>
<hr />
<p><span class="math-container">$(3)$</span> Let <span class="math-container">$u^{2} = t+1$</span>.</p>
<p><span class="math-container">$(4)$</span> Let <span class="math-container">$v^{2} = 1-t$</span> and let <span class="math-container">$w^{2} = t-1$</span>.</p>
<hr />
<p>Putting everything together, we have</p>
<p><span class="math-container">$$ \begin{align} \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx &= - \frac{\sqrt{\pi}}{2} \Im \left(e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc}(1) + \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right) \right)\\ &= - \frac{\sqrt{\pi}}{2} \left(\frac{e \sqrt{\pi}}{\sqrt{2}} \operatorname{erfc}(1) - \frac{\sqrt{\pi} }{\sqrt{2}e} \operatorname{erfi}(1) - \frac{\sqrt{\pi}}{\sqrt{2}e} \right) \\ &= \frac{\pi}{2 \sqrt{2}e} \left(1+ \operatorname{\color{red}{erfi}}(1) - e^{2} \operatorname{erfc}(1) \right). \end{align}$$</span></p>
|
63,534 | <p>I found in an article <a href="http://dx.doi.org/10.1103/PhysRev.105.776" rel="nofollow">"Imperfect Bose Gas with Hard-Sphere Interaction"</a>, <em>Phys. Rev.</em> 105, 776–784 (1957) the following integral, but I don't know how to solve it. Any hints?</p>
<p>$$\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\frac{\sinh(upq)}{q^2 - p^2}pq} } e^{-vq^2 - wp^2} = \frac{\pi}{4}\frac{u(w - v)}{\left[(w + v)^2-u^2 \right]\left(4wv-u^2\right)^{1/2}}$$</p>
<p>for $u,v,w > 0$.</p>
| Andrew | 11,265 | <p>First of all one can note that the integral converges and is a differentiable function of parameters for $4vw>u^2$. With the change of variables $p\to p' u^{1/2}$, $q\to q' u^{1/2}$ the general case be reduced to $u=1$. Now denoting the lhs
$$
f(v,w)=\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\frac{\sinh(pq)}{q^2 - p^2}pq} } e^{-vq^2 - wp^2}
$$
we have
$$
\frac{\partial}{\partial v}f(v,w)-\frac{\partial}{\partial w}f(v,w)=
\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\sinh(pq)pq}} e^{-vq^2 - wp^2}
=\frac{\pi }{2 (4 v w-1)^{3/2}},
$$
the integral converging for $vw>1/4$.
Since $f(w,v)=-f(v,w)$ we have $f(v,v)=0\;$. The solution of this Cauchy problem can be obtained in the standard way (rotating the coordinate system on $\pi/4$ etc.):<br>
$$
f(v,w)=\frac{\pi (w-v)}{4 \left((v+w)^2-1\right)\sqrt{4 v
w-1} }\;.
$$</p>
|
13,467 | <p>I will present two problems alongside solutions, student is doing problems of type I like a cakewalk but has several issues with the problems of type 2;</p>
<p><strong>Type I</strong></p>
<blockquote>
<p>Consider an experiment of rolling two dice:</p>
<p>Sample space $$ S = \{(1,1),(1,2),(1,3), \cdots, (6,6) \} $$</p>
<p>Let $A$ be the event of getting 6 as sum on two dice:</p>
<p>Event $$A = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$ </p>
<p>Let $B$ be the event of getting 4 on first die:</p>
<p>Event $$B = \{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\}$$ </p>
<p>Now, the probability of getting sum of 6 on two dice given that first
die appears as 4 is given by</p>
<p>$$p(A \mid B) = \dfrac{p(A\cap B)}{p(B)} = \dfrac{p\{(4,2)\}}{p(B)} =
\dfrac{1}{6}$$</p>
</blockquote>
<p><strong>Type II</strong></p>
<blockquote>
<p>Let $C$ be the event of having cancer and $p(C) = 1/2$</p>
<p>The probability of having both tumor and cancer is $p(C\cap T) = 1/6$</p>
<p>Then the probability of having tumor given cancer is given by</p>
<p>$$p(T\mid C) = \dfrac{p(T\cap C)}{p(C)} = \dfrac{1}{3}$$</p>
</blockquote>
<p>Students are understanding type I problems and type 2 problems well, but some students are asking to present $C, T, C\cap T$ in terms of sets. I tried to convince them using Venn diagrams. But they are asking for either roster form (for discrete sets) or set builder form(for any set).</p>
<p>I am trying to do like follows </p>
<p>\begin{align}
\text{Sample space} = S &= \{x \mid \text{$x$ is a living being} \} \\
C &= \{x \mid \text{$x$ has cancer}\} \\
C\cap T &= \{x \mid \text{$x$ has both tumor and cancer}\} \\
\end{align}</p>
<p>Is it right way to do? Students are facing difficulty and they are asking every event inform of set, since definition of event is that it is a subset of sample space (which is a set).</p>
| Andreas Blass | 326 | <p>Since they can do Type-1 problems, I'd begin by trying to make your Type-2 problem look as similar to Type 1 as possible, despite the fact that, in the Type-2 problem, we don't actually know the size of the sample space. So I'd temporarily assume some specific size, say $6$ (to make the arithmetic easy). So we have $6$ people of whom $3$ have cancer, and one of those $3$ also has a tumor. Let them work out the conditional probability in this case, which I hope they can do because this version of the question is very similar to Type 1. Then, re-do the problem with a sample space of size, say, 24. And again with 60 or 600, etc., until it becomes clear to them that the information in the problem is enough to determine the answer, regardless of the size of the sample space. The formula for conditional probability in terms of absolute probabilities summarizes the results of these calculations.</p>
|
3,879,366 | <p>This might or might not be author specific, but in Algebraic Statistics by Seth Sullivant the author uses a notation of <span class="math-container">$S_{K, K}$</span> for example. So far I have not been able to find out what it means. Here is an <a href="https://i.stack.imgur.com/oEJHw.png" rel="nofollow noreferrer">example</a>. For reference, <a href="https://i.stack.imgur.com/43Wn6.png" rel="nofollow noreferrer">here</a> is the syntax the author uses for an induced subgraphs.</p>
| Zac | 521,980 | <p>I haven't solved it yet but figured that the function is reasonably inspectable via plotting in this case. Some simple python code allows us to visually identify where the minimum should be, although from the answers so far no analytical solution appears forthcoming:</p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
g = norm.cdf
x = np.linspace(-1,1)
# Objective func to maximise:
f = x*(1 - g(x) - (g(x))**2)
# Cost func to minimise:
L = -f
plt.plot(x,L)
plt.grid()
</code></pre>
<p>From this, the minimum appears to be around 0.015. <a href="https://i.stack.imgur.com/fsM1b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fsM1b.png" alt="The plot produced by the above code" /></a></p>
|
4,141,812 | <p>Considering the vector space <span class="math-container">$\mathbb{R}^3$</span>, find all values of <span class="math-container">$k$</span> such that
<span class="math-container">\begin{align*}
\begin{bmatrix}
2k^2 \\
-3k \\
1
\end{bmatrix} \in \text{ span }
\left\{
\begin{bmatrix}
1 \\
1 \\
3
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix},
\begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
\right\}.
\end{align*}</span></p>
<p>Here is my approach:<br />
By creating the system of equations
<span class="math-container">\begin{align*}
\begin{cases}
x_1 + x_3 &= 2k^2\\
x_1 + x_2 + 2x_3 &= -3k\\
3x_1 + x_2 + 4x_3 &= 1
\end{cases}
\end{align*}</span>
And then, I acquired its RREF by performing elementary row operations.
<span class="math-container">\begin{align*}
\begin{bmatrix}
1 & 0 & 1 & 2k^2 \\
0 & 1 & 1 & -2k^2-3k \\
0 & 0 & 0 & -4k^2 + 3k + 1
\end{bmatrix}
\end{align*}</span>
Since we get a pivot point at the augmented column,
<span class="math-container">\begin{align*}
-4k^2+3k+1 &= 0\\
k &= 1, -\dfrac{1}{4}
\end{align*}</span>
I'm not really sure with what I did in the last part. Is this right?</p>
| Ritam_Dasgupta | 925,091 | <p>Yes, the vertical asymptote would occur at some <span class="math-container">$x=a$</span> such that: <span class="math-container">$$e^a=-a$$</span>
I do not think there is any elementary way to find this <span class="math-container">$a$</span>, but you could plot it and find out.</p>
|
692,383 | <p>Let $p, q$ be prime numbers which may or may not be distinct.
Let $\mathbb{Q}_p$ be the field of $p$-adic numbers.
Let $\mathbb{Z}_p$ be the ring of $p$-adic integers.
We define similarly $\mathbb{Q}_q$ and $\mathbb{Z}_q$.</p>
<p>Let $A = \mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$.
Let $\lambda\colon \mathbb{Z}_p \rightarrow A$
and $\mu\colon \mathbb{Z}_q \rightarrow A$ be the canonical ring homomorphisms.
Let $B = \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$.
Let $\psi\colon B \rightarrow A$ be the ring homomorphism induced by $\lambda$ and $\mu$.</p>
<p>Is $\psi$ injective?</p>
| Makoto Kato | 28,422 | <p>All rings will be assumed to be commutative with unity.</p>
<p><strong>Lemma 1</strong>
Let $A$ be a ring, $M, N$ $A$-modules.
Suppose $N$ is flat over $A$.
Let $a$ be an element of $A$.
Suppose the map $f\colon M \rightarrow M$ defined by $f(x) = ax$ is injective.
Then the map $g\colon M\otimes_A N \rightarrow M\otimes_A N$ defined by $g(z) = az$ is also injective.</p>
<p>Proof:
Let $1_M\colon M \rightarrow M$ be the identity map.
Since $ax\otimes y = a(x\otimes y)$ for $x \in M, y \in N$, $g = f\otimes 1_M$.
Hence $g$ is injective.</p>
<p><strong>Lemma 2</strong>
Let $A$ be a ring, $N$ an $A$-module.
Let $(M_i)_{i\in I}$ a <a href="http://en.wikipedia.org/wiki/Direct_limit" rel="nofollow noreferrer">direct system</a> of $A$-modules.
Then colim$_i (M_i\otimes_A N) =$ (colim$_i M_i)\otimes_A N$.</p>
<p>Proof:
This is well known.
See, for example, Matsumura's Commutative Ring Theory, Appendix A1.</p>
<p><strong>Remark</strong>
Lemma 2 can be proved by using category theory as follows.
$-\otimes_A N$ is a left adjoint functor of Hom$_A(N, -$)(see my answer to <a href="https://math.stackexchange.com/questions/18881/motivation-for-tensor-product">this question</a>).
Hence it commutes with colim(see MacLane: Categories for the working mathematician, Chapter V, Section 5, Theorem 1, p.114).</p>
<p><strong>Lemma 3</strong>
Let $A$ be a ring.
Let $(M_i)_{i\in I} \rightarrow (N_i)_{i\in I}$
and $(N_i)_{i\in I} \rightarrow (L_i)_{i\in I}$ be maps of direct sytems of $A$-modules.
Suppose each sequence $M_i \rightarrow N_i \rightarrow L_i$ is exact.
Then colim $M_i \rightarrow$ colim $N_i \rightarrow$ colim $L_i$ is exact.</p>
<p>This is well known.
See, for example, Matsumura's Commutative Ring Theory, Appendix A2.</p>
<p><strong>Lemma 4</strong>
Let $A$ be a ring.
Let $(M_i)_{i\in I}$ a direct system of flat $A$-modules.
Then colim $M_i$ is flat.</p>
<p>Proof:
This follows immediately from Lemma 2 and Lemma 3.</p>
<p><strong>Lemma 5</strong>
Let $G$ be a torsion-free abelian group.
In other words, the order of every nonzero element of $G$ is infinite.
Then $G$ is flat.
In other words, the functor $-\otimes_{\mathbb{Z}} G$ is exact.</p>
<p>Proof:
Let $(G_i)_{i \in I}$ be the family of finitely generated subgroups of $G$.
Since each $G_i$ is torsion-free, it is free.
Hence it is flat.
Hence $G =$ colim $G_i$ is flat by Lemma 4.</p>
<p><strong>Proposition</strong>
Let $p, q$ be prime numbers which may or may not be distinct.
Let $A = \mathbb{Q}_p\otimes_{\mathbb{Q}} \mathbb{Q}_q$.
Let $\lambda\colon \mathbb{Z}_p \rightarrow A$
and $\mu\colon \mathbb{Z}_q \rightarrow A$ be the canonical ring homomorphisms.
Let $B = \mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$.
Let $\psi\colon B \rightarrow A$ be the ring homomorphism induced by $\lambda$ and $\mu$.
Then $\psi$ injective.</p>
<p>Proof:
Let $S = \mathbb{Z} - \{0\}$.
Then $S^{-1} (\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q) \cong \mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_q$ by Lemma 3 of my answer to <a href="https://math.stackexchange.com/questions/684146/is-mathbbz-p-otimes-mathbbz-mathbbz-q-noetherian">this question</a>.
Since $\mathbb{Z}_q$ is torsion-free, it is flat by Lemma 5.
Hence $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$ is torsion free by Lemma 1.
Hence $\psi$ injective.</p>
|
712 | <p>I read <a href="https://math.stackexchange.com/questions/625/why-is-the-derivative-of-a-circles-area-its-perimeter-and-similarly-for-spheres">this question</a> the other day and it got me thinking: the area of a circle is $\pi r^2$, which differentiates to $2 \pi r$, which is just the perimeter of the circle. </p>
<blockquote>
<p>Why doesn't the same thing happen for squares? </p>
</blockquote>
<p>If we start with the area formula for squares, $l^2$, this differentiates to $2l$ which is sort of right but only <em>half</em> the perimeter. I asked my calculus teacher and he couldn't tell me why. Can anyone explain???</p>
| user76995 | 76,995 | <p>The area of a circle can be defined as a summation of infinite concentric circles which are infinitely small, each of radius $r$ perimeter $2\pi r$ where $r$ is a variable that changes from $0$ to $R$, where $R$ denotes the radius of the circle we defined.</p>
<p>So, obviously when the area of the circle is differentiated, you will find $dA$ equal to $2\pi rdr$ (It's not $2 \pi R$).</p>
|
47,728 | <p>Function <code>CreateTemporary</code> can generate a unique file name like</p>
<pre><code>file = CreateTemporary[]
(* "C:\Documents and Settings\kkouptsov\Local \Settings\Temp\m-263a0380-3a03-49ac-95cb-d21390b2d3fd" *)
</code></pre>
<p>How do we generate a unique file name such as above example? The reason I don't use <code>CreateTemporary</code> because I don't need to actually create the file, but only need the file path.</p>
<p>I plan to generate many of these names in parallel, and they should be unique to each other.</p>
| Sascha | 4,597 | <p>Version 10.0 introduced <code>CreateUUID</code></p>
<pre><code>CreateUUID[]
(* "1aaeab6f-ad51-48d2-a939-366da47374a3" *)
</code></pre>
<p>The <a href="http://reference.wolfram.com/language/ref/CreateUUID.html" rel="nofollow noreferrer">documentation</a> says that</p>
<blockquote>
<ul>
<li><p>CreateUUID[] gives a UUID based on a 128-bit number, formatted as 32 hexadecimal digits grouped in the format 8-4-4-4-12.</p>
</li>
<li><p>CreateUUID[] makes use of detailed local system and session information, as well as absolute time.</p>
</li>
<li><p>Some Wolfram System features depend on assuming the UUIDs will not collide.</p>
</li>
</ul>
</blockquote>
<p>Just for fun I tested this with 100.000 UUIDs</p>
<pre><code>Table[CreateUUID[], 100000] // Apply@UnsameQ
(* True *)
</code></pre>
|
941,596 | <p>Assuming I'm flipping $M$ biased coins with different probability for heads $p_i, i=\{1,...,M\}$. What is the probability of having $k$ times head? Is there a distribution function known for this?</p>
| Ri-Li | 152,715 | <p>There is no known distribution function but you can create by your own.
Order the biased coin $1,2 \ldots ,M$. Now in the binomial coefficient we have $\binom Mk \times p^k \times q^{M-k}$ . Here for a fixed $k$ it will be $ (p_{i_1}p_{i_2}p_{i_3} \ldots p_{i_k})(q_{i_1}q_{i_2}q_{i_3} \ldots q_{i_k})$ product will run over all possible choosing of $k$ in $M$. Now add them. Now construct the distribution.</p>
|
83,236 | <p>I need a <code>compare-operation</code> for two lists of same length (usually > 100000) which does the following:</p>
<blockquote>
<p>{2,3,5,4,1,8,7} <code>compare-operation</code> {1,4,6,3,2,8,8} = {2,4,6,4,2,8,8}</p>
</blockquote>
<p>The resulting list has at each position the greater (or equal element in case both are equal) of the two lists (because 2>1,4>3,6>5,4>3,2>1,8=8,8>7).</p>
<p>How can that be done?</p>
| LLlAMnYP | 26,956 | <pre><code>MapThread[Max, {{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}}]
</code></pre>
<blockquote>
<p><code>{2, 4, 6, 4, 2, 8, 8}</code></p>
</blockquote>
<pre><code>Thread[Unevaluated[Max[{2, 3, 5, 4, 1, 8, 7} , {1, 4, 6, 3, 2, 8, 8}]]]
</code></pre>
<p>also works. <code>Unevaluated</code> is necessary, otherwise <code>Max[...</code> evaluates, before <code>Thread</code> acts on it. On the other hand</p>
<pre><code>Thread[Unevaluated[Max[list1, list2]]]
</code></pre>
<p><em>does not work</em> for the same reasons.</p>
<pre><code>list1 = RandomInteger[30, 1*^7];
list2 = RandomInteger[30, 1*^7];
MapThread[Max, {list1, list2}] // AbsoluteTiming
</code></pre>
<blockquote>
<p><code>{6.302683, {28, 10, 21, 13 ...</code></p>
</blockquote>
<p>Not too bad for ten million elements. In fact, faster than constructing the lists.</p>
<p>In this simple case however, we can do <strong>10 times better</strong>:</p>
<pre><code>(Max /@ Transpose@{list1, list2}) // AbsoluteTiming // First
</code></pre>
<blockquote>
<p><code>0.608412</code></p>
</blockquote>
<p>For a list with 2 dimensions the obvious solution is</p>
<pre><code>MapThread[Max, {list1, list2}, 2] // AbsoluteTiming // First
</code></pre>
<blockquote>
<p><code>0.702000</code></p>
</blockquote>
<p>Timing is shown for a <code>10^3</code> by <code>10^3</code> list.</p>
<p>The <em>faster</em> solution uses the, so-to-speak, generalized transpose, which is achieved by using flatten with a matrix as the second argument:</p>
<pre><code>Map[Max, Flatten[{list1, list2}, {{2}, {3}, {1}}], {2}] // AbsoluteTiming // First
</code></pre>
<blockquote>
<p><code>0.109200</code></p>
</blockquote>
<p>As before, almost an order of magnitude faster. The general expression for <code>k</code>-dimensional lists would be:</p>
<pre><code>Map[Max, Flatten[{list1, list2}, {{2}, {3}, ..., {k+1}, {1}}], {k}]
</code></pre>
<p><strong>Update 18.09.17</strong></p>
<p>Carl Woll's <code>ThreadedMax</code> is hard to beat, but exploiting <code>Compile</code>d listability comes close:</p>
<pre><code>threadMax =
Compile[{{x1, _Integer}, {x2, _Integer}}, Max[x1, x2],
RuntimeAttributes -> {Listable}, Parallelization -> True,
CompilationTarget -> "C", RuntimeOptions -> "Speed"]
</code></pre>
<p>It is still a few times slower though and I'd recommend Carl's solution. Simple top-level functions with built-in listability are close to impossible to beat even with compiled functions and are more versatile in the datatypes they accept: e.g., the above <code>threadMax</code> will only work on packed arrays of integers.</p>
<p>If <code>Ramp</code> isn't available in your version of MMA, just like it is in mine (I'm on 10.2), I suggest the following implementation:</p>
<pre><code>ThreadedMax[l1_List, l2_List] :=
With[{diff = Subtract[l1, l2]},
Check[UnitStep[diff] diff + l2, $Failed]]
</code></pre>
<p>I would similarly modify Carl's solution like so for better performance:</p>
<pre><code>ThreadedMax[l1_List, l2_List] := Check[
Ramp[Subtract[l1, l2]]+l2,
$Failed
]
</code></pre>
<p>See <a href="https://mathematica.stackexchange.com/q/40927/26956">this</a> for further reading.</p>
|
83,236 | <p>I need a <code>compare-operation</code> for two lists of same length (usually > 100000) which does the following:</p>
<blockquote>
<p>{2,3,5,4,1,8,7} <code>compare-operation</code> {1,4,6,3,2,8,8} = {2,4,6,4,2,8,8}</p>
</blockquote>
<p>The resulting list has at each position the greater (or equal element in case both are equal) of the two lists (because 2>1,4>3,6>5,4>3,2>1,8=8,8>7).</p>
<p>How can that be done?</p>
| Carl Woll | 45,431 | <p>For threading the maximum of a list (or array) with a number, you'll be hard pressed to beat the following:</p>
<pre><code>ThreadedMax[l1_List, l2_?NumericQ] := Clip[l1, {l2, Infinity}]
</code></pre>
<p>For example:</p>
<pre><code>l1 = RandomReal[1, 10^8];
ThreadedMax[l1, .5]; //AbsoluteTiming
</code></pre>
<blockquote>
<p>{0.552601, Null}</p>
</blockquote>
<p>For threading the maximum of 2 arrays, the following is the fastest approach I can think of:</p>
<pre><code>ThreadedMax[l1_List, l2_List] := Check[
Ramp[l1-l2]+l2,
$Failed
]
</code></pre>
<p>Comparing with the answer by @LLlAMnYP:</p>
<pre><code>l1 = RandomReal[1, 10^7];
l2 = RandomReal[1, 10^7];
r1 = ThreadedMax[l1, l2]; //AbsoluteTiming
r2 = MapThread[Max, {l1, l2}]; //AbsoluteTiming
r3 = Max /@ Transpose[{l1, l2}]; //AbsoluteTiming
r1 === r2 === r3
</code></pre>
<blockquote>
<p>{0.089001, Null}</p>
<p>{5.53821, Null}</p>
<p>{0.602344, Null}</p>
<p>True</p>
</blockquote>
<p>For arrays with higher rank, we have:</p>
<pre><code>l1 = RandomReal[1, {3000, 3000}];
l2 = RandomReal[1, {3000, 3000}];
r1 = ThreadedMax[l1, l2]; //AbsoluteTiming
r2 = MapThread[Max, {l1, l2}, 2]; //AbsoluteTiming
r3 = Map[Max, Flatten[{l1, l2}, {{2}, {3}, {1}}], {2}]; //AbsoluteTiming
r1 === r2 === r3
</code></pre>
<blockquote>
<p>{0.096501, Null}</p>
<p>{5.30103, Null}</p>
<p>{0.694005, Null}</p>
<p>True</p>
</blockquote>
|
4,340,560 | <p>I have to teach limits to infinity of real functions of one variable.
I would like to start my course with a beautiful example, not simply a basic function like <span class="math-container">$1/x.$</span> For instance, I thought of using the functions linked to the propagation of covid-19 and show that, under the basic model, the number of contaminations will go to <span class="math-container">$0$</span> when time goes to <span class="math-container">$+\infty.$</span> However, this is a bad idea because the model is not so easy to explain and moreover students are sick of covid-subjects.</p>
<p>Hence, I ask you some help to find interesting examples from physics, geography, etc ... I suppose that an example with "time" going to <span class="math-container">$+\infty$</span> would be nice.</p>
| Exponent BOT | 404,974 | <p>Explaining the failure of Zeno's Paradox could be a cool activity. Motion can be broken down into so many very small parts, namely steps of length <span class="math-container">$1/2^i$</span>. Show the fact that <span class="math-container">$\lim_{n \to \infty} \sum_{i=1}^n 1/(2_i) =1$</span>. The only prerequisite of this is understanding geometric series.</p>
<p>I would also suggest compound interest and its limiting case of continuous compounding, as a previous poster suggested.</p>
|
3,009,351 | <p>Consider the vectors <span class="math-container">$q_1=(1,1,1)$</span> and <span class="math-container">$q_3=(1,1,-2)$</span>. I need to find a third vector <span class="math-container">$q_2$</span> such that <span class="math-container">$\{q_1,q_2,q_3\}$</span> is a arthogonal basis for <span class="math-container">$\mathbb{R}^3$</span>. </p>
<p>My problem is the following: I did take <span class="math-container">$v=(1,0,0)$</span> and I did verify that <span class="math-container">$\{q_1,q_3,v\}$</span> is a basis for <span class="math-container">$\mathbb{R}^3$</span>. Then I did take <span class="math-container">$$q_2=v-\langle v|q_1\rangle q_1-\langle v|q_3\rangle q_3=(-1,-2,1)$$</span></p>
<p>And, by Gram-Schmidt process, <span class="math-container">$q_2$</span> must be orthogonal to <span class="math-container">$q_1$</span> and <span class="math-container">$q_3$</span>. But, as we can see, it does not happen. So, where is my mistake?</p>
| Siong Thye Goh | 306,553 | <p><span class="math-container">$$q_2=v-\frac{\langle v|q_1\rangle}{\langle q_1|q_1\rangle} q_1-\frac{\langle v|q_3\rangle}{\langle q_3|q_3\rangle} q_3$$</span></p>
|
3,387,843 | <p>The thing I want to know is that why the term <span class="math-container">$``vector"$</span>.</p>
<p>I had learned that vector is something that has magnitude and direction, and also the the elements of <span class="math-container">$\mathbb{R}^n$</span> i.e. <span class="math-container">$n$</span> tuples can be visualized as vectors . We can also perform vector addition and scalar multiplication over them.
So this was fine.
But again there are also vector spaces where I am not able to visualize the elements as the vectors like.</p>
<p>The set of <span class="math-container">$n\times n$</span> matrices whose elements are chosen from the set of real numbers forms a vector space over <span class="math-container">$\mathbb{R}$</span>.
Also the set of real valued continuous functions forms vector space.</p>
<p>So are this elements vectors or am I mistaken with the meaning of vectors?</p>
| user | 505,767 | <p>The ordinary geometric vectors in <span class="math-container">$\mathbb R^2$</span> or in <span class="math-container">$\mathbb R^3$</span> are a special kind of a more general structure named <a href="https://en.m.wikipedia.org/wiki/Vector_space" rel="nofollow noreferrer">vector space</a> which obeys to the same rules and that we can extend to others object such as matrices, polynomials, etc. which can be added together and multiplied by scalar from any field (usually <span class="math-container">$\mathbb R$</span> or <span class="math-container">$\mathbb C$</span>).</p>
|
3,154,947 | <p>Hello MathStackExchange Community, </p>
<p>Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations. </p>
<ul>
<li><em>The generic formula is : <span class="math-container">$P(n,r)= n!/(n-r)!$</span></em></li>
<li><span class="math-container">$n!$</span> = being the total number of objects in an option set</li>
<li><span class="math-container">$(n-r)!$</span> = the limitations caused by the choices made.</li>
</ul>
<p><em>but this only works for very simple permutation-processes where you make one single choice-type.</em></p>
<p>If I have 15 particles, 10 neutral and 5 positive particles
and I care about the arrangement</p>
<p>he formula is <span class="math-container">$P(n,r)= 15!/10!\cdot5!$</span><br>
<strong>What I don't get is how does 10! 5! translate to <span class="math-container">$(n-r)!$</span></strong></p>
<p><strong>the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?</strong></p>
<p><em>What is the way to think about the second portion?</em> Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.</p>
<p>Thanks, </p>
<p>ThermoRestart </p>
| B. Goddard | 362,009 | <p>I think this is a Green's theorem problem. The loop is traced as <span class="math-container">$t$</span> goes from <span class="math-container">$-4$</span> to <span class="math-container">$4$</span> (but clockwise.) So the area is</p>
<p><span class="math-container">$$-\frac{1}{2} \int_{-4}^4 x \; dy - y \; dx $$</span> <span class="math-container">$$ =
-\frac{1}{2} \int_{-4}^{4} (49-t^2)(3t^2-16)-(t^3-16t)(-2t) \; dt $$</span> <span class="math-container">$$= \frac{8192}{15} = 546.13\ldots.$$</span></p>
<p>The extra minus sign is because of the clockwise orientation.</p>
|
475,267 | <p>Suppose $f(0) =0 $ and $0<f''(x)<\infty (\forall$ $x>0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases. </p>
<p>I have shown that $f'(x)-\frac{f(x)}{x} = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here? </p>
| Pete L. Clark | 299 | <p>The hypothesis <span class="math-container">$f'' \geq 0$</span> means that your function is <strong>convex</strong>. There are many standard inequalities involving convex functions. This is a special case of one of them, which I call the <strong>two secant inequality</strong>:</p>
<blockquote>
<p>If <span class="math-container">$f$</span> is convex and <span class="math-container">$a < x < b$</span>, <span class="math-container">$\frac{f(x)-f(a)}{x-a} \leq \frac{ f(b)-f(a)}{b-a}$</span>.</p>
</blockquote>
<p>(Since your function satisfies <span class="math-container">$f'' > 0$</span> it is <strong>strictly convex</strong>, and then this and other related inequalities can be taken to be strict.)</p>
<p>I discuss this inequality in <span class="math-container">$\S$</span> 7.3.4 <a href="http://alpha.math.uga.edu/%7Epete/2400full.pdf" rel="nofollow noreferrer">of these notes</a>. Unfortunately the slightly later proof that <span class="math-container">$f'' \geq 0$</span> implies this inequality is -- as I've just noticed -- faulty. Instead of the above inequality I derive the weaker inequality <span class="math-container">$\frac{f(x)-f(a)}{x-a} \leq \frac{f(b)-f(x)}{b-x}$</span>. (In fact convex functions satisfy the <strong>three-secant inequality</strong> <span class="math-container">$\frac{f(x)-f(a)}{x-a} \leq \frac{ f(b)-f(a)}{b-a} \leq \frac{f(b)-f(x)}{b-x}$</span>. Please draw a picture!)</p>
<p>So I feel fortunate that Ted Shifrin has sketched an alternate proof. Let's flesh it out. Assuming <span class="math-container">$a = f(a) = 0$</span> and <span class="math-container">$f'' > 0$</span>, we must show that the function <span class="math-container">$g(x) = \frac{f(x)}{x}$</span> is strictly increasing on <span class="math-container">$(0,\infty)$</span>. Its derivative is</p>
<p><span class="math-container">$g'(x) = \frac{xf'(x) - f(x)}{x^2}$</span>,</p>
<p>so we want <span class="math-container">$h(x) = xf'(x) - f(x) > 0$</span> for all positive <span class="math-container">$x$</span>; then <span class="math-container">$g' > 0$</span>, so <span class="math-container">$g$</span> is strictly increasing. Since <span class="math-container">$h(0) = 0$</span>, it will be enough to show that <span class="math-container">$h'(x) > 0$</span> for all <span class="math-container">$x > 0$</span>. And indeed</p>
<p><span class="math-container">$h'(x) = f'(x) + xf''(x) - f'(x) = xf''(x) > 0$</span> for all <span class="math-container">$x > 0$</span>.</p>
<p>I will fix the proof of Theorem 7.18 in my notes when I get the chance. (<b>Added</b>: I incorporated into my notes the proof of Ted Shifrin in the comment below, which uses only that <span class="math-container">$f'$</span> is increasing. This is with respect to my private comment. I am also writing notes on the sequential completion of an ordered field which -- surprise, surprise -- contains some annoying details, so I haven't uploaded the corrected copy yet but I should do so relatively soon.)</p>
|
2,307,399 | <blockquote>
<p>How to solve $$\sum_{i=1}^n \lfloor e\cdot i \rfloor $$
For a given $n$. </p>
</blockquote>
<p>For example, if $n=3$, then the answer is $15$, and it's doable by hand. But for larger $n$ (Such as $10^{1000}$) it gets complicated . Is there a way to calculate this summation?</p>
| infinitylord | 178,643 | <p><strong>disclaimer:</strong> This is not a complete answer, just an approximation I came up with that was too long to comment.</p>
<p>If we truncate $e $ at $3$ decimal places (so it is 2.718), we can get a decent approximation:</p>
<p>$$\sum_{i=1}^n \lfloor 2.718i \rfloor \approx \frac32 (n(n+1) - \lfloor \frac{n}3 \rfloor ( \lfloor \frac{n}3 \rfloor + 1)) + \frac12 \lfloor \frac{n}4 \rfloor ( \lfloor \frac{n}4 \rfloor + 1) = f(n)$$</p>
<p>I used the program provided in the comments to compare the answers for $n = 542$ (I know, not very large, but the program timed out for me there).</p>
<p>$$\sum_{k=1}^{542} \lfloor ei \rfloor = 399,732$$</p>
<p>$$f(542) = 401,769$$</p>
<p>So that the error is roughly $0.51$%</p>
<p>From here, you could numerically find an approximation for the error as a function of $n$, say $\delta(n)$, and a much better approximation for large values would be $f(n) - \delta(n) $ </p>
<p>You would never find an exact answer this way, but if you're only curious in the <em>value</em> the sum takes for large $n$, this could be a decent way to get a nice approximation (it could be hard for a code to compute a sums exact value at very large $n $)</p>
|
2,307,399 | <blockquote>
<p>How to solve $$\sum_{i=1}^n \lfloor e\cdot i \rfloor $$
For a given $n$. </p>
</blockquote>
<p>For example, if $n=3$, then the answer is $15$, and it's doable by hand. But for larger $n$ (Such as $10^{1000}$) it gets complicated . Is there a way to calculate this summation?</p>
| Matthew Conroy | 2,937 | <p>Let's approximate $e$ by a rational number $A$ such that we have a positive integer $k$ such that $10^kA$ is an integer (e.g., $A=2.71$ with $k=2$).</p>
<p>We can then use the fact that $\lfloor (m\cdot10^k+i)A \rfloor = m\cdot 10^k A+\lfloor Ai \rfloor$ for any positive integer $m$.</p>
<p>Then, for a positive integer $r$,
$$ \large
\sum_{i=1}^{r \cdot 10^k} \lfloor Ai \rfloor = r \sum_{i=1}^{10^k} \lfloor Ai \rfloor +\frac{r(r-1)}{2}10^{2k}A.$$</p>
<p>Since the sum is a non-decreasing function of $A$, we can bound the desired sum with two
rational approximations of $e$, one large and one smaller. </p>
<p>For instance, with $A=2.71828$ and $A=2.71829$ (and $k=5$, $r=10^4$) we can find that
$$
1359140000859160000 < \sum_{i=1}^{10^9} \lfloor ei \rfloor < 1359145000859150000
$$
With $A=2.7182818$ and $A=2.7182819$ (and $k=7$, $r=10^2$) we find that
$$
1359140900859141000 <\sum_{i=1}^{10^9} \lfloor ei \rfloor < 1359140950859140900.$$
By using better approximations, we can get tighter bounds.</p>
|
2,297,542 | <blockquote>
<p>Given integers $a,b,c,d$ with $d\not\equiv0\bmod5$ and $m$ an integer for which $am^3+b m^2+cm+d\equiv0\bmod5$, prove that there exists an integer $n$ for which $dn^3+cn^2+bn+a\equiv0\bmod5$.</p>
</blockquote>
| Landon Carter | 136,523 | <p>Let $Y\sim Exp(1)$ then $-Y$ has the desired cdf.</p>
|
3,469,392 | <p>Let <span class="math-container">$a_n$</span> be a recursive sequence where <span class="math-container">$a_1=42$</span> and for all <span class="math-container">$n$</span>:</p>
<p><span class="math-container">$$a_{n+1}=a_n+\frac{(-1)^n(1+n!)}{2^nn!}$$</span></p>
<p>After plugging some values of <span class="math-container">$n$</span> I can "see" that the sequence is bounded at <span class="math-container">$(41,42]$</span></p>
<p>And that for <span class="math-container">$a_{2n}$</span> the sequence is increasing and for <span class="math-container">$a_{2n+1}$</span> it is decrecasing</p>
<p>I know that I need to show that the sequence is either monotonic decrecasing <span class="math-container">$(a_n\geq a_{n+1})$</span> or monotonic increasing <span class="math-container">$(a_n\leq a_{n+1})$</span></p>
<p>I have tried to set <span class="math-container">$a_{n+1}=a_n=L$</span> too, but it did not work, how should I approach this?</p>
<p>One way that I know is to look at <span class="math-container">$\frac{a_{n+1}}{a_{n}}$</span> and to see if it <span class="math-container">$\leq 1$</span> or <span class="math-container">$\geq 1$</span> </p>
| Robert Z | 299,698 | <p>We have that
<span class="math-container">$$a_{N+1}=a_1+\sum_{n=1}^N\frac{(-1)^n(1+n!)}{2^nn!}=42+
\sum_{n=1}^N\frac{(-1/2)^n}{n!}+\sum_{n=1}^N\left(-\frac{1}{2}\right)^n$$</span>
and as <span class="math-container">$N\to +\infty$</span> both series can be easily evaluated: the first one is related to the <a href="https://en.wikipedia.org/wiki/Exponential_function#Formal_definition" rel="noreferrer">power series of <span class="math-container">$e^x$</span></a> and the second one is a <a href="https://en.wikipedia.org/wiki/Geometric_series#Formula" rel="noreferrer">geometric series</a>.</p>
|
432,223 | <p>I wanted to know, how can I factor $x^6 +5x^3 +8$, I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do.</p>
<p>Help appreciated.</p>
<p>Thanks.</p>
| Glad Galaxy | 241,926 | <p>I think here for factoring the polynomial $x^6+5x^3+8$ we must go ahead by the same method of Eric Tressler. But, when we get the roots, it is seen that two complex roots are conjugate and so form the factor $x^2-x+2$. Hence, the factorization of Veritas is obtained. I think with this way Veritas get the factorization.</p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| martin | 13,496 | <p>To give a German-centric answer which would certianly apply to other countries as well: </p>
<p>The German Abitur is the right to study any subject at a university (Allgemeine Hochschulreife). As such the pupils need to be taught everything such that they need in order to be able to succeed. After all, they may want to change the subject after finishing school and you do not want them to have to decide at an early age.</p>
<p>In other countries, a similar concept exists at different levels. As such, a broad education is necessary. Which is why me, as a mathematician, needed to study history and social studies, which I clearly did not have any interest in. </p>
|
17,885 | <p>In most education systems, Mathematics is a compulsory subject from primary school all the way to the start of university. A common reason given is that essential concepts like addition and multiplication are taught to the children. </p>
<p>But for many high school students, especially those who are keen on pursuing the humanities, they do not see any point in studying Mathematics for the rest of their schooling life, or how any concept in Mathematics could possibly be applied in their future work.</p>
<p>Why is Mathematics a compulsory subject for high school students, especially those who are clearly studying in Humanities streams?</p>
| Russell McMahon | 13,518 | <p>We are allowing people with less than even basic numeracy to leave school.<br>
We are allowing people to enter university with not only no knowledge of basic algebra or calculus but also with no realisation that basic algebra and calculus are essential tools when dealing with subjects such as history or social sciences. </p>
<p>A social scientist or historian needs a basic understanding of exponential growth or decay - if only enough to provide a platform for more learning when the need latterly becomes obvious. Higher levels of knowledge can be far more easily be built on a basic foundation than being obtained with no prior introduction. </p>
<p>Consider:</p>
<p>You buy 5 items that cost 0.30 currency units each.<br>
You give the shop assistant a 5 currency unit note.<br>
They pick up a calculator to multiply 0.3 by 5 and then to subtract this answer from 5.<br>
They give you 3 currency units in change.<br>
They wonder why you look at them in what used to be astonishment until it happened many many many times. </p>
<p>That's just the most basically basic of numeracy.<br>
Now for some basic mind bending with algebra and geometry ...</p>
<p>If we are releasing people into the world who cannot calculate 5 - 0.3 * 5 (as we are) then we are stopping far short of a reasonable target. </p>
<p>I have seen social science papers which are utter unmitigated and irredeemable balderdash BECAUSE the writer has no grasp of basic mathematical concepts, and doesn't know that they don't know</p>
<p>STAY AT SCHOOL*.</p>
<hr>
<p>*Failing to gain a minimum broad education is functionally failing to 'stay at school' </p>
|
3,043,699 | <p>I'm having difficulty understanding how below integral is evaluated : </p>
<p><span class="math-container">$$\int_0^1\frac{(1-y)^2}{2}+(1-y)y dy = \frac{-1(1-y)^3}{6}+\frac{y^2}{2}-\frac{y^3}{3}\bigg|^1_0$$</span></p>
<p>What are steps involved in this evaluation ?</p>
<p>For <span class="math-container">$\frac{(1-y)^2}{2}$</span> this appears to be evaluated as <span class="math-container">$\frac{(1-y)^{2+1}}{(3)2}$</span> but I'm how unsure - 1 is inserted ?</p>
| Community | -1 | <p>By the chain rule,</p>
<p><span class="math-container">$$((ax+b)^k)'=(ax+b)'k(ax+b)^{k-1}=ak(ax+b)^k$$</span></p>
<p>and from this,</p>
<p><span class="math-container">$$\int(ax+b)^kdx=\frac{(ax+b)^{k+1}}{a(k+1)}.$$</span></p>
<p>Try with <span class="math-container">$a=-1$</span>.</p>
|
2,534,369 | <p>I am trying to work my through the exercises in Spivak's <em>Calculus on Manifolds.</em> I am currently working on the exercises in Chapter 3 which deals with Integration. I am having trouble with the following question:</p>
<blockquote>
<p>Let:</p>
<p>\begin{equation}
f(x,y)=\begin{cases}
0, & \text{if $x$ is irrational}.\\
0, & \text{if $x$ is rational, $y$ is irrational}. \\
1/q, & \text{if $x$ is rational, $y=p/q$ in lowest terms}.
\end{cases}
\end{equation}</p>
<p>Show that $f$ is integrable on $A = [0,1] \times [0,1]$ and $\int_A f = 0$.</p>
</blockquote>
<p>I was thinking of trying to prove that this set is Jordan Measurable and that it's Jordan measure is zero and that it is therefore Riemann Integrable but I am not sure how to do this or if it is even the best way to solve this problem.</p>
<p>If I could show that $f$ is continuous on $A$ up to a set of Jordan Measure $0$, then $f$ would be integrable but again, I'm not sure I can do this or if its even appropriate for this problem.</p>
<p>Any assistance that anyone could provide would be greatly appreciated.</p>
<p>Thank you.</p>
| RAUL PEREZ | 963,490 | <p>It is obvious that the inferior sums are zero, so we want to proof that the superior sums <span class="math-container">$U(f;P)$</span> can be taken less than any previously fixed <span class="math-container">$\epsilon$</span>. Let p prime and let <span class="math-container">$k=\max \{n \mid n(n+1)/2 +1\leq p)\}$</span> . Then <span class="math-container">$p=\frac{k(k+1)}{2}+q$</span>, with <span class="math-container">$0\leq q\leq k+1$</span>. (If <span class="math-container">$q=0$</span> then <span class="math-container">$p$</span> is of the form <span class="math-container">$p=1+2++...+k=k(k+1)/2$</span> and if <span class="math-container">$q=k+1$</span>, <span class="math-container">$p $</span> is of the same form <span class="math-container">$p=1+2+...+k+k+1=(k+1)(k+2)/2$</span>). Now we take the partition <span class="math-container">$P=\{0<1/p<...<(p-1)/p<1\}$</span>. The reason for taking <span class="math-container">$p$</span> prime is that any fraction <span class="math-container">$a/b\in [\frac{m-1}{p},\frac{m}{p} ]$</span> with <span class="math-container">$b<p$</span> is the interior of the interval. That avoid that the same fraction repeats when we take the supreme of the function. For example, if <span class="math-container">$p=4$</span>, then <span class="math-container">$1/2$</span> appears as the supreme value of <span class="math-container">$f$</span> in <span class="math-container">$[1/4,1/2]$</span> and <span class="math-container">$[1/2,3/4]$</span> and we don't want this to happen. Thus, <span class="math-container">$1$</span> appears only one time and no more as the supreme of the upper interval <span class="math-container">$[\frac{p-1}{p},1]$</span>.
<span class="math-container">$1/2$</span> appears also only one time and no more as the supreme on some interval for the reason explained before. For <span class="math-container">$1/3$</span> we have that it can appear maybe two times and no more as the supreme of <span class="math-container">$f$</span> on the intervals containing <span class="math-container">$1/3$</span> and <span class="math-container">$2/3$</span> and <span class="math-container">$1/4$</span> may appear <span class="math-container">$3$</span> times and more as the supreme of <span class="math-container">$f$</span> on some intervals (in fact <span class="math-container">$1/4$</span> is the supreme of <span class="math-container">$f$</span> in at most 2 intervals: one containing <span class="math-container">$1/4$</span> and the other containing <span class="math-container">$3/4$</span> because <span class="math-container">$2/4=1/2$</span>, but we want to obtain an upper bound of <span class="math-container">$U(f; P)$</span> so we will count <span class="math-container">$1/4$</span> three times). We can go on with this reasoning and we will have that as upper bounds of the supremes we have <span class="math-container">$1$</span> for one interval, <span class="math-container">$1/2$</span> also for one interval, <span class="math-container">$1/3$</span> for <span class="math-container">$2$</span>, <span class="math-container">$1/4$</span> for <span class="math-container">$3$</span>,... and <span class="math-container">$1/(k+1)$</span> for <span class="math-container">$k $</span> intervals. Until now we have <span class="math-container">$1+1+2+3+...+k=1+\frac{k(k+1)}{2}$</span> intervals for which we have found upper bounds of the supreme of <span class="math-container">$f$</span> on them. For the remaining <span class="math-container">$q$</span> intervals we can choose <span class="math-container">$1/(k+2)$</span> as an upper bound. So, we have
<span class="math-container">$$U(f;P)\leq \frac{1}{p}[1+1/2+2/3+3/4+...+k/(k+1)+q/(k+2)]=\frac{1}{p}[1+1-1/2+1-1/3+1-1/4+...+1-1/(k+1)+q/(k+2)]=\frac{1}{p}[k+1-S+q/(k+2)]$$</span>, where <span class="math-container">$S=1/2+...+1/(k+1)>0$</span>. Then
<span class="math-container">$$U(f;P)<\frac{1}{p}[k+1+q/(k+2)]$$</span>
But <span class="math-container">$1/p<\frac{2}{k(k+1)}$</span>, since <span class="math-container">$p>\frac{k(k+1)}{2}$</span>, so
<span class="math-container">$$U(f;P)<\frac{2}{k(k+1)}[k+1+q/(k+2)]=\frac{2}{k}+\frac{2q}{k(k+1)(k+2)}\leq \frac{2}{k}+\frac{2}{k(k+2)}$$</span> since <span class="math-container">$q\leq k+1$</span>.
The last expression tends to <span class="math-container">$0$</span> as <span class="math-container">$k$</span> tends to infinite. But <span class="math-container">$p \longrightarrow \infty \Rightarrow k\longrightarrow \infty$</span> so <span class="math-container">$$U(f;P)\longrightarrow 0$$</span> if <span class="math-container">$p \longrightarrow \infty$</span></p>
|
123,675 | <p>What are some "applications to" / "connections with" topology that one could hope to reasonably cover in a first course on topos theory (for master students)? I have an idea of what parts of the theory I would like to cover, however, I would love some more nice examples and applications. Of course, I have some ideas of my own, but am open to suggestions. Thank you!</p>
<p>P.S.</p>
<p>I am also interested in perhaps learning about some new connections myself, which would be out of reach for such a course, so feel free to leave these as well, qualified as such.</p>
| Mike Shulman | 49 | <p>This is probably too advanced for a first course, but you might be interested in Ieke Moerdijk's monograph "Classifying Spaces and Classifying Topoi". Among other things, it uses topos theory to show what it is that the "classifying space" of a (non-groupoid) category classifies. IIRC this was an ingredient in the proof of the Madsen-Weiss theorem.</p>
<p>More obviously, one could say something about sheaf cohomology.</p>
|
2,008,437 | <p>Given four points in the plane, there exists a one-dimensional family of conics through these, often called a pencil of conics. The locus of the centers of symmetry for all of these conics is again a conic. What's the most elegant way of computing it?</p>
<p>I know I could choose five arbitrary elements from the pencil, compute their centers and then take the conic defined by these. I can also do so on a symbolic level, to obtain a general formula. But that formula is at the coordinate level, and my CAS is still struggeling with the size of the polynomials involved here. There has to be a better way.</p>
<p>Bonus points if you know a name for this conic. Or – as the center is the pole of the line at infinity – a name for the more general locus of the pole of an arbitrary line with respect to a given pencil of conics.</p>
| Blue | 409 | <p>Here's a coordinate derivation that's computationally intense, although my copy of Mathematica was able to crunch the symbols fairly quickly.</p>
<hr>
<p>Let the points $P$, $Q$, $R$, $S$ be arranged such that segments $\overline{PR}$ and $\overline{QS}$ meet at the origin, and such that the coordinate axes bisect the angles created by those segments. (This will give us a nice symbolic symmetry in $x$ and $y$.) Thus, we may assign
$$\begin{gather}
P = p\;(\phantom{-}\cos\theta, \phantom{-}\sin\theta) \quad Q = q\;(\phantom{-}\cos\theta, -\sin\theta) \\
R = r\;(-\cos\theta, -\sin\theta) \quad S = s\;(-\cos\theta,\phantom{-}\sin\theta )\end{gather}$$</p>
<p>(So, for positive $p$, $q$, $r$, $s$, we have opposite rays $\overrightarrow{OP}$ and $\overrightarrow{OR}$, and opposite rays $\overrightarrow{OQ}$ and $\overrightarrow{OS}$, emanating from the origin.)</p>
<p>We'll parameterize the one-parameter pencil of conics by conveniently parameterizing the variable fifth point that defines the members of the pencil:
$$T := t\; ( P + Q )$$</p>
<p><em>Mathematica</em> tells me that the corresponding conic's equation ...
$$a x^2 + b x y + c y^2 + d x + e y + f = 0$$
... has these coefficients:
$$\begin{align}
a &= -\phantom{2}(r s + (q r + p s - 2 r s) t + (p - q )(r-s) t^2) \sin^2\theta \\
b &= \phantom{-} 2 (p r - q s) t^2 \sin\theta \cos\theta\\
c &= \phantom{-2}(r s + (q r + p s - 2 r s) t - ( p + q )( r + s ) t^2) \cos^2\theta \\
d &= \phantom{-} 2 (p q r + p q s - p r s - q r s) t^2 \sin^2\theta \cos\theta\\
e &= - 2 (p q r - p q s - p r s + q r s) t^2 \sin\theta\cos^2\theta \\
f &= \phantom{-} 4 p q r s t^2 \sin^2\theta \cos^2\theta
\end{align}$$</p>
<p>Now, referring to <a href="https://math.stackexchange.com/a/801572/409">this tremendously-convenient answer</a>, we have that the conic's center has coordinates</p>
<p>$$x = \frac{b e - 2 c d}{4 a c - b^2} \qquad y = \frac{b d - 2 a e}{4 a c - b^2}$$</p>
<p>The numerators and denominators of these expressions are slightly-hairy fourth-degree polynomials in $t$ that I won't transcribe. From here, I invoked <em>Mathematica</em>'s <code>Resultant[]</code> function to eliminate the parameter $t$; the result(ant), after ignoring extraneous factors, is:</p>
<blockquote>
<p>$$\begin{align}
0 \quad=\quad &\phantom{2} x^2 ( p r - q s )\sin^2\theta \\
+\; &\phantom{2} y^2 ( p r - q s ) \cos^2\theta \\
-\; &2 x y ( p r + q s ) \sin\theta \cos\theta \\
+\; &\phantom{2} x ( (p-r) qs - pr(q-s) ) \sin^2\theta \cos\theta \\
+\; &\phantom{2} y ( (p-r) qs + pr(q-s) ) \sin\theta \cos^2\theta
\end{align} \tag{$\star$}$$ </p>
</blockquote>
<p>With no constant term, $(\star)$ describes a conic through the origin; that is, <strong>the conic-of-centers passes through the intersection of $\overline{PR}$ and $\overline{QS}$</strong>. Of course, if the conic passes through the intersection of <em>one</em> pair of segments determined by the original points, then it must pass through the intersections of <em>the other two pairs</em> of segments, as well. (Symbolic verification is left to the reader.) Thus, three points of the conic-of-centers are easily determined geometrically ... and they might even be "obvious" in a projective approach.</p>
<p>The all-important discriminant has the remarkably-simple form:</p>
<blockquote>
<p>$$16 p q r s \sin^2\theta \cos^2\theta$$</p>
</blockquote>
<p>from which we can say</p>
<blockquote>
<p>$$\text{the conic-of-centers is} \begin{cases}
\text{an ellipse} &\text{if}\;pqrs < 0 \\
\text{a parabola} &\text{if}\;pqrs = 0 \\
\text{an hyperbola} &\text{if}\;pqrs > 0 \end{cases}
$$</p>
</blockquote>
<p>The eccentricity computation is messy, so I'll leave that to the reader. On the other hand, the center is quite nice: it is, in fact, the centroid of $P$, $Q$, $R$, $S$:</p>
<blockquote>
<p>$$\text{center of the conic-of-centers}\; = \frac{1}{4}( P + Q + R + S )$$</p>
</blockquote>
<p>I guess I'll stop here.</p>
<hr>
<p>One more thing ... If $p$, $q$, $r$, $s$ are all non-zero, then we can define reciprocals $\overline{p} := 1/p$, etc, and divide $(\star)$ through by $pqrs$ to get this form of the equation:</p>
<blockquote>
<p>$$\begin{align}
0 \quad&=\phantom{2} x^2 ( \overline{p}\overline{r} - \overline{q}\overline{s} )\sin^2\theta \\
&+\phantom{2} y^2 ( \overline{p}\overline{r} - \overline{q}\overline{s} ) \cos^2\theta \\
&+2 x y ( \overline{p}\overline{r} + \overline{q}\overline{s} ) \sin\theta \cos\theta \\
&+ \phantom{2} x ( (\overline{p}-\overline{r}) - (\overline{q} - \overline{s}) ) \sin^2\theta \cos\theta \\
&+ \phantom{2} y ( (\overline{p}-\overline{r}) + (\overline{q} - \overline{s} )) \sin\theta \cos^2\theta
\end{align} \tag{$\overline{\star}$}$$</p>
</blockquote>
|
1,085,164 | <p><strong>I am currently taking a course on Numerical PDE. The course covers the following topics listed below.</strong> </p>
<p><strong>Chapter 1: Solutions to Partial Dierential Equations:</strong></p>
<p><strong>Chapter 2: Introduction to Finite Elements:</strong></p>
| dustin | 78,317 | <p>A book I like, but I am probably biased since he was my professor is Dr. Jianke Yang's book <a href="http://rads.stackoverflow.com/amzn/click/0898717051" rel="nofollow">Nonlinear Waves in Integrable and Non-integrable systems</a></p>
<p>It covers:</p>
<ol>
<li>Derivation of non-linear waves</li>
<li>Integrable theory for the non-linear Schrodinger equation</li>
<li>Theories for integrable equations with order scattering operators</li>
<li>Soliton perturbation theory and applications</li>
<li>Theories for non-integrable systems</li>
<li>Nonlinear waves phenomena in periodic media</li>
<li>Numerical methods for non-linear wave equations</li>
</ol>
<p>Also, Matlab code is provided in the book.</p>
|
2,315,599 | <p>I am trying to evaluate the integral<br>
$\int (\frac{1}{z(\exp(z)-1)}dz$ $\mathbb{C}:|z|=1\}$</p>
<p>using Laurent theorem. $0$ is a singularity point and both denominators becomes $0$ when $z=0$. So it's hard to find a method to evaluate this using Residue theorem. Could someone please show me how to solve this.
Thank you.</p>
| Johnq | 233,719 | <p>Soumee,</p>
<p>I believe I may have an answer that is founded in equations based off your idea from before.</p>
<p>Set up a reference frame with left denoting the negative direction and right being positive. Place the girl's house at the origin. Using the 1-D kinematics equations for position, let the following being true:</p>
<p>a) $x_s = -(v_s)*t$ Where $x_s$ is Sania's position on the line and $v_s$ is the velocity of Sania.</p>
<p>b) $x_t = (v_t)*t - x_{0,t}$ Where $x_t$ is the position of the train on<br>
the line and $v_t$ is the velocity of the train and $x_{0,t}$ is the
train's position at the time Sania usually leaves her house.</p>
<p>If you let the two positions equal each other and solve for $x_{0,t}$, you get an equation analogous to your equation 1 from before. In the form of my equations this is:</p>
<p>c) $x_{0,t} = (v_s + v_t)(t_{M})$</p>
<p>Where $t_{M}$ is the usual meet up time</p>
<p>Next, set them equal again, but this time set the new meet up time to be 3 minutes behind $t_{M}$. The following equations should come:</p>
<p>d)</p>
<p>$$-(v_s)*(t_{M}-3 min)-(12 min * v_s) = (v_t)*(t_{M}-3 min) - x_{0,t}$$</p>
<p>This equality comes from the fact that they meet at the same position on the 1-D reference frame and for the fact that at t=0, Sania has already traveled for 12 minutes. Solving the equation for $v_t$ gives you the speed of the train. It should also be noted that both sides will contain equation c, so you can go ahead and cancel them on both sides of the second equality.</p>
<p>CHANGELOG:---------</p>
<p>CHANGE 1: ACCOUNTING FOR SIGN ERROR MADE (negative sign was changed to positive for $v_t$ so Sania and the train go in opposite directions)</p>
<p>CHANGE 2: ACCOUNTING FOR ERROR IN POSITION (equation d was reconfigured so that the train keeps its original position at time t=0 and accounts for the fact that for 12 minutes before t=0 , Sania was heading in the left direction).</p>
<p>CHANGE3: ADDED CLARIFICATION ON EQUATION d (elaborated on why the origin term $-(12 min * v_s)$ was put in and elaborated more on the choice for time (here t=0 is where we start the "clock" so to speak. Sania has already traveled a certain distance toward the meeting point at the time the train hits the point on the line where it usually is when Sania usually leaves).</p>
|
423,302 | <p>I have a perfectly fair coin, and my goal is to prove that it is unfair with a confidence level of 95%. In order to accomplish this, <em>I will cheat</em>. Whenever I fail to have enough evidence, I will simply increase the sample size by continuing to flip the coin.</p>
<p>To be specific, I will flip the coin until the proportion of heads is either small enough or large enough to be able to say that the coin is unfair. Let's say that, for any given sample size $N$, there is at least a $95\%$ chance that the proportion of heads is within the confidence interval</p>
<p>$$0.5 ± f(N)$$</p>
<p>where $f(N)$ is some function. If I have flipped the coin a grand total of $N$ times, and the proportion of heads is outside of this interval, then I "conclude" that the fair coin is unfair and stop the process. If the proportion is within the interval, then I flip the coin one more time and repeat the process with $N+1$. One important detail is that I never "throw out" data.</p>
<p>What is the expected number of flips that I would have to make until I receive the "statistically significant" result I seek?</p>
<p>I've run a few simulations with my calculator, and it seems that either it takes a reasonable number of flips (like 44) or it takes a <em>huge</em> amount of time. Why is this, assuming it's not programmer error?</p>
<h2>Edit: More simulations</h2>
<p>I ran some more simulations (on an actual PC, rather than a calculator). I used two criteria for determining whether or not a result is significant. First, I required that there be at least $20$ flips performed. Second, I used this value for the function:</p>
<p>$$f(N) = 1.9600 * \sqrt{\frac{0.5 * 0.5}N} = \frac{0.98}{\sqrt{N}}$$</p>
<p>More edits: this formula comes from the normal approximation for the binomial distribution, where $1.96$ is the required z-score and other part is the standard deviation of the proportion.</p>
<p>My plan for running the simulations was to set an upper limit for the number of flips, preform a huge number of trials, and see what proportion of trials exceeded the limit. This way, I could gather data in a reasonable time limit. After running a large number of simulations, I acquired this data:</p>
<pre><code># of flips proportion > this number sample size
100000 0.298 n=1000
10000 0.41 n=1000
1000 0.5516 n=10000
100 0.7637 n=10000
32 0.8726 n=10000
64 0.8029 n=10000
128 0.738 n=10000
256 0.6768 n=10000
512 0.6252 n=10000
1024 0.563 n=10000
2048 0.5126 n=5000
</code></pre>
| not all wrong | 37,268 | <p><strong>Edited Note</strong>: From David Speyer's answer the below reaches the correct conclusion -- divergence in the expectation value -- though I haven't verified the intermediate observation about the actual asymptotics. (There was a bit of confusion at one point.) </p>
<hr>
<p>I ran 10000000 simulations (up to <strong>100000 max tosses each</strong>) while watching TV and collected a histogram of where I stopped. I got slightly different numbers to you, but sufficiently similar that I think this is a combination of statistical errors, rounding errors and coding errors! I suspect the basic behaviour is the same, but it's possible there is some extreme sensitivity in the tails. I haven't checked my code for bugs at all.</p>
<hr>
<p>Haven't thought about why this behaviour arises, but it for $M$ the number of tosses required, the probability <strong>looks like</strong> it behaves as
$$\mathbb{P}(M \ge m) \sim \frac C {m^k}\qquad C\approx 1.41,k\approx 0.13$$
<img src="https://i.stack.imgur.com/IujmQ.png" alt="Graph"></p>
<p>Of course, this might just be a completely misleading line, it's very difficult to tell. This is definitely empirical.</p>
<p>If this holds up to further scrutiny, then we certainly have
$$\mathbb{E}(M)=\sum_m^\infty \mathbb P(M\ge m) = \infty$$
If someone else fancies a bit of fun, I think seeing this is the correct behaviour and deriving an expression for $k$ might be a nice challenge!</p>
<hr>
<p><strong>Edit</strong>: Updated with slightly better data.</p>
<p>Also, as I said in the comment above,</p>
<blockquote>
<p>This is the expected stopping time of a random walk Markov chain with moving barriers; since random walks move $\sim \sqrt N$ and your barriers are moving at (roughly the same) $\sim \sqrt N$ it's possible the expectation diverges.</p>
</blockquote>
<p>You can probably come up with a decent argument as to why this sort of behaviour is to be expected, but I haven't taken taken the time to try it yet, so that might be overly optimistic. If I get round to thinking about this I'll post!</p>
|
423,302 | <p>I have a perfectly fair coin, and my goal is to prove that it is unfair with a confidence level of 95%. In order to accomplish this, <em>I will cheat</em>. Whenever I fail to have enough evidence, I will simply increase the sample size by continuing to flip the coin.</p>
<p>To be specific, I will flip the coin until the proportion of heads is either small enough or large enough to be able to say that the coin is unfair. Let's say that, for any given sample size $N$, there is at least a $95\%$ chance that the proportion of heads is within the confidence interval</p>
<p>$$0.5 ± f(N)$$</p>
<p>where $f(N)$ is some function. If I have flipped the coin a grand total of $N$ times, and the proportion of heads is outside of this interval, then I "conclude" that the fair coin is unfair and stop the process. If the proportion is within the interval, then I flip the coin one more time and repeat the process with $N+1$. One important detail is that I never "throw out" data.</p>
<p>What is the expected number of flips that I would have to make until I receive the "statistically significant" result I seek?</p>
<p>I've run a few simulations with my calculator, and it seems that either it takes a reasonable number of flips (like 44) or it takes a <em>huge</em> amount of time. Why is this, assuming it's not programmer error?</p>
<h2>Edit: More simulations</h2>
<p>I ran some more simulations (on an actual PC, rather than a calculator). I used two criteria for determining whether or not a result is significant. First, I required that there be at least $20$ flips performed. Second, I used this value for the function:</p>
<p>$$f(N) = 1.9600 * \sqrt{\frac{0.5 * 0.5}N} = \frac{0.98}{\sqrt{N}}$$</p>
<p>More edits: this formula comes from the normal approximation for the binomial distribution, where $1.96$ is the required z-score and other part is the standard deviation of the proportion.</p>
<p>My plan for running the simulations was to set an upper limit for the number of flips, preform a huge number of trials, and see what proportion of trials exceeded the limit. This way, I could gather data in a reasonable time limit. After running a large number of simulations, I acquired this data:</p>
<pre><code># of flips proportion > this number sample size
100000 0.298 n=1000
10000 0.41 n=1000
1000 0.5516 n=10000
100 0.7637 n=10000
32 0.8726 n=10000
64 0.8029 n=10000
128 0.738 n=10000
256 0.6768 n=10000
512 0.6252 n=10000
1024 0.563 n=10000
2048 0.5126 n=5000
</code></pre>
| David E Speyer | 448 | <p>According to Blackwell and Freedman, <a href="http://projecteuclid.org/euclid.aoms/1177703292" rel="nofollow">A Remark on the Coin Tossing Game</a>, the expected time for the number of heads to get outside $n/2 \pm a \sqrt{n}$ is finite for $a<0.5$ and infinite for $a \geq 0.5$. (As Henry points out below, they use $\pm 1$ instead of $(0,1)$ counting, so their boundary is $0 \pm \sqrt{n}$.) I'm afraid I don't understand their proof, but this means that the expectation time should be infinite, as Sharkos finds.</p>
|
2,036,278 | <p>I have a question about inverse images and containment. </p>
<p>Let $\;f: A \rightarrow B, D \subseteq A$, and $E \subseteq B$. Show that $\; f^{-1}(B-E) \subseteq A-f^{-1}(E)$. </p>
<p>I have started by letting $x \in f^{-1}(B-E)$ so $f(x) \in (B-E)$, but I am unsure where to go from here.</p>
| Olivier Oloa | 118,798 | <p>Setting
$$
S_N:=\sum_{l=2}^N{\frac{1}{l^m}}, \quad m \ge 3
$$ we have
$$
S_{N+1}-S_N=\frac{1}{l^m}>0,
$$
$\{S_N\}$ is an increasing sequence then observing that, for $m\ge3$,
$$
S_N=\sum_{l=2}^N{\frac{1}{l^m}}<\sum_{l=2}^N{\frac{1}{l(l+1)}}=\frac12-\frac1{N+1}<\frac12
$$ $\{S_N\}$ is then a bounded sequence, the given series is thus convergent.</p>
|
4,263,836 | <blockquote>
<p>I would like to compute <span class="math-container">$\mathrm{Ext}^i_{\mathbb Z}(\mathbb Q, \mathbb Z/2\mathbb Z)$</span>.</p>
</blockquote>
<p>So from the definition I learned in my class, I need to find a free resolution of <span class="math-container">$\mathbb Q$</span> over <span class="math-container">$\mathbb Z$</span>, the first step is to find a surjection of <span class="math-container">$P^0\to\mathbb Q$</span> where <span class="math-container">$P^0$</span> is a free <span class="math-container">$\mathbb Z$</span>-module. Then apply <span class="math-container">$\text{Hom}_\mathbb Z(-,\mathbb Z/2\mathbb Z)$</span> and compute the cohomology. However, I find it hard to come up with an easy surjection onto <span class="math-container">$\mathbb Q$</span>. Since <span class="math-container">$\mathbb Q$</span> is countable, we can find a bijection <span class="math-container">$\phi:\mathbb Z\to\mathbb Q$</span> and consider the infinite direct sum <span class="math-container">$\oplus_{i=1}^{\infty} \mathbb Z$</span> and a map <span class="math-container">$$\psi:\oplus_{i=1}^{\infty}\mathbb Z\to \mathbb Q$$</span> given by <span class="math-container">$\psi((a_1, a_2,\ldots))=\sum_{i=1}^\infty a_i\phi(i).$</span> This is clearly a surjection and a morphism. However, it is hard to compute the cohomology coming from this map since I do not know how to characterize its kernel.</p>
<p>So I am just wondering is there a "better" resolution which can helps me compute Ext functor easily? Any hint is appreciated.</p>
| Thomas Andrews | 7,933 | <p>If you must do it with a free resolution, consider the free group <span class="math-container">$\oplus_{i=1}^\infty \mathbb Z$</span> and morphism:</p>
<p><span class="math-container">$$\phi:(a_i)_{i=1}^\infty \mapsto \sum_{i=1}^\infty \frac{a_i}{i!}$$</span></p>
<p>Then the kernel has as a basis <span class="math-container">$e_n=(e_{ni})_i$</span> with <span class="math-container">$$e_{ni}=\begin{cases}1&i=n\\-(n+1)&i=n+1\\0&\text{otherwise}\end{cases}$$</span>
Then: <span class="math-container">$$\phi(e_n)=\frac1{n!}-\frac{n+1}{(n+1)!}=0.$$</span></p>
<hr />
<p>Showing this is a free basis is not hard.</p>
<p>The <span class="math-container">$e_n$</span> are independent because if <span class="math-container">$b=(b_i)=\sum c_ne_n$</span>, then for the smallest <span class="math-container">$k$</span> such that <span class="math-container">$c_k\neq 0,$</span> then <span class="math-container">$b_k=c_k\neq0,$</span> so <span class="math-container">$b\neq0.$</span></p>
<p>Showing they generate the kernel is only a little harder.</p>
<p>If <span class="math-container">$a=(a_i)$</span> is a non-zero element in the kernel, let <span class="math-container">$$m(a)=\min\{i\mid a_i\neq0\}\\n(a)=\max\{i\mid a_i\neq 0\}.$$</span></p>
<p>If <span class="math-container">$m(a)=n(a),$</span> then <span class="math-container">$a$</span> can’t be non-zero, because <span class="math-container">$\phi(a)=\frac{a_m}{m!},$</span> and <span class="math-container">$a_m\neq0.$</span></p>
<p>So <span class="math-container">$m(a)<n(a).$</span></p>
<p>We can define <span class="math-container">$$a’=a-a_{m(a)}e_{m(a)}.$$</span></p>
<p>Then <span class="math-container">$a’$</span> is in the kernel, and if non-zero, then <span class="math-container">$m(a’)>m(a)$</span> and <span class="math-container">$n(a’)=n(a).$</span></p>
<p>So by induction on <span class="math-container">$n(a)-m(a),$</span> you eventually get to zero, and the original element was a combination of the <span class="math-container">$e_m.$</span></p>
|
3,676,127 | <p>Is there a term in mathematics for the <strong>SET</strong> of all possible <span class="math-container">$n \times n$</span> zero-one matrices?</p>
<p>In other words, for a particular <span class="math-container">$n \in \mathbb{Z}^+$</span>, is there a term for the set of all possible <span class="math-container">$M=[m_{ij}]$</span> such that either <span class="math-container">$m_{ij}=0$</span> or <span class="math-container">$m_{ij}=1$</span> for <span class="math-container">$1\le i \le n$</span> and <span class="math-container">$1\le j \le n$</span>?</p>
<p><span class="math-container">$\{ M=[m_{ij}] : m_{ij}=1 \vee m_{ij}=0, 1\le i \le n, 1\le j \le n\}$</span></p>
<p>Thank you.</p>
| J.-E. Pin | 89,374 | <p>It is the set <span class="math-container">${\Bbb B}^{n \times n}$</span> of <span class="math-container">$n \times n$</span> Boolean matrices. You can extend the notation to <span class="math-container">${\Bbb B}^{n \times m}$</span>, the set of <span class="math-container">$n \times m$</span>-matrices with entries in the Boolean semiring.</p>
|
1,842,093 | <p><a href="https://i.stack.imgur.com/A5U3D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A5U3D.png" alt="enter image description here"></a></p>
<p>$ABCD$ forms a square. $CDE$ forms a triangle. Given $\measuredangle AED=15^{\circ}$ and $DE=CE$, prove $\triangle CDE$ is equilateral. </p>
<p>The question is surprising hard, the problem is basically proving $\measuredangle BAE=\measuredangle AEB$. Can I have some hint. </p>
| Nathaniel B | 204,043 | <p>Each possible length of $DE$ both is uniquely determined by and uniquely determines angle $\measuredangle CDE$. Likewise, $\measuredangle DEA$ is uniquely determined by and uniquely determines either one of the length $DE$ or $\measuredangle CDE$. This means that we can reverse the problem! Thus "Prove that if triangle $CDE$ is equilateral, $\measuredangle DEA=15^{\circ}$" is an equivalent problem. This is much easier. Proof:</p>
<p>1) Triangle $DEA$ is isoceles (since $DE=AD$)</p>
<p>2)$\measuredangle ADE=(90+60)^{\circ}=150^{\circ}$</p>
<p>3)$\measuredangle DEA=((180-150)/2)^{\circ}=15^{\circ}$</p>
<p>And we are done.</p>
|
2,976,057 | <p>I’m doing partial fractions and need to factorize the denominator. They are quadratic. However there are some that aren’t so easy to factorize and my first choice was to use the quadratic equation to find the roots however comparing my answer with the correct one the signs are different. Is the quadratic formula only to be used when the equation is equal to zero? The answer used another method of factorizing that didn’t involve equating anything to zero and I can’t find anything about it online. Where did I go wrong?
My denominator is: </p>
<p><span class="math-container">$-3z^2 -4z-1$</span></p>
<p>the correct answer is:
<span class="math-container">$-(3z+1)(z+1)$</span></p>
<p>while if I do this using the quadratic formula I get:
<span class="math-container">$(3z+1)(z+1)$</span></p>
<p>however if I factorize the negative sign then use the quadratic formula I get the correct answer which is confusing to me.</p>
| Joppy | 431,940 | <p>Suppose I have a mystery quadratic <span class="math-container">$p(x) = -2x^2 + x + 1$</span>, and want to factorise it using the quadratic formula. I can find the roots easily enough:</p>
<p><span class="math-container">$$ x = \frac{-1 \pm \sqrt{1 + 8}}{-4}$$</span>
and so the roots are <span class="math-container">$-\frac{1}{2}$</span> and <span class="math-container">$1$</span>, and so I <em>think</em> that the polynomial should factorise as <span class="math-container">$p(x) = (x - 1)(x+\frac{1}{2})$</span>. But this is wrong, since I forgot that <span class="math-container">$p(x)$</span> and <span class="math-container">$2p(x)$</span> and <span class="math-container">$\frac{-p(x)}{9}$</span> and so on all have the same roots! And so knowing the two roots only determines <span class="math-container">$p(x)$</span> up to scaling. But it's easy to find the right scaling factor, by just looking at the <span class="math-container">$-2x^2$</span> term. So I arrive at the <em>actual</em> answer, of
<span class="math-container">$$ p(x) = -2x^2 + x + 1 = -2(x-1)(x + \frac{1}{2})$$</span></p>
<p>So after you find the roots, remember to multiply the whole thing by the right number to fix up the <span class="math-container">$x^2$</span> term.</p>
|
663,218 | <p>Basic question in linear algebra here. $T$ is a linear transform from $\mathbb R^n$ to $\mathbb R^n$ defined by $T(v)=Av$, $A\in \mathrm{Mat}_n(\mathbb R)$. We are given some inner product $\langle ,\rangle$ of $\mathbb R^n$. Does not have to be the standard one, just some random inner product.</p>
<p>let $T^*$ be a linear transform from $\mathbb R^n$ to $\mathbb R^n$ such that for all $u,v \in \mathbb R^n$: $\langle T(u),v\rangle=\langle u,T^*(v)\rangle$</p>
<p>I know that if $\langle ,\rangle$ is the standard inner product of $\mathbb R^n$, then $T^*(v)=A^tv$. My question is, does this hold for all inner products?</p>
<p>If it isn't, given some inner product and the transform $T$, how can I find $T^*$?</p>
<p>And if it is true, then why?</p>
| Yiorgos S. Smyrlis | 57,021 | <p>No it does not hold for any inner product.</p>
<p>It is hard hard to show that any other inner product $\langle\cdot,\cdot\rangle_*$ can be represented as
$$
\langle x,y\rangle_\star=\langle x,Sy\rangle,
$$
where $S$ is a positive definite matrix. So
$$
\langle Tx,y\rangle_\star=\langle x,T^*Sy\rangle \ne \langle x,ST^*y\rangle=\langle x,T^*y\rangle_\star,
$$
unless $ST^*=T^*S$.</p>
|
2,027,529 | <p>I want to show that $\pi_1(S^1\vee S^1) = \mathbb{Z}*\mathbb{Z}. $ I know that it follows from the Seifert-van Kampen theorem, but we haven't talked about that in class. </p>
<p>We are given the following hint:</p>
<blockquote>
<p>Define a group homomorphism $\mathbb{Z}*\mathbb{Z}\to\pi_1(S^1\vee S^1)$ via the two inclusions $S^1\to S^1\vee S^1,$ and show that this is an isomorphism. </p>
</blockquote>
<p>We've also never talked about free products, so I am very confused as to how to define this homomorphism.</p>
| Ben Grossmann | 81,360 | <p>Present the group $\Bbb Z * \Bbb Z = \langle a,b \rangle$. It suffices to define our map on the generators $a,b$.</p>
<p>Let $L(S^1 \vee S^1)$ denote the group of loops over composition, and $\sim$ be the homotopy equivalence relation so that $\pi_1(X) = L(X)/\sim$. Let $f$ be a map that sends $a$ to a loop around the first copy of $S^1$ send $b$ to a loop around the second copy of $S^1$. Our homomorphism is $\pi_\sim \circ f$.</p>
<p>The trick is to show that this is an isomorphism. How you do so depends on your tool set. It's likely you have a theorem which looks something like this:</p>
<blockquote>
<p><strong>Theorem 59.1:</strong> Suppose $X = U \cup V$, where $U,V$ are open sets of $X$. Suppose that $U \cap V$ is path connected, and that $x_0 \in U \cap V$. Let $i$ and $j$ be the inclusion mappings of $U$ and $V$ respectively into $X$. Then the images of the induced homomorphisms
$$
i_*: \pi_1(U,x_0) \to \pi_1(X,x_0) \qquad
j_*: \pi_1(V,x_0) \to \pi_2(X,x_0)
$$
generate $\pi_1(X,x_0)$ <em>(Munkres, "Topology" (Second Edition))</em></p>
</blockquote>
<p>As it applies to our problem: the key is that every loop through $x_0$ (the intersection of our two circles) can be written as the composition of finitely many loops that "simple loops", i.e. those that begin and end at $x_0$ without passing through it again. With that, we'll see that the homomorphism is surjective.</p>
<p>not sure about injectivity, though.</p>
|
1,620,795 | <p>Find function f(x), where:
$$f(3)=3$$
$$f'(3)=3$$
$$f'(4)=4$$
$$f''(3) = \nexists$$ </p>
<p>How to find function like this in <strong>general?</strong> What steps should I do?</p>
| TZakrevskiy | 77,314 | <p>$$|x\cdot y| \le \sum_{i=1}^n |x_i\bar y_i|\le \sum_{i}\left((\sup_j|x_j||y_i|)\right)=\sup_j|x_j|\sum_i|y_i|=\|x\|_{\infty}\|y\|_1$$</p>
|
607,353 | <p>So I've recently been looking at the <a href="http://www.photographyblogger.net/stereographic-projection-photography-100-little-planets/" rel="nofollow">Tiny Planet</a> images.</p>
<p>I've been googling a few things to try and find out how images are converted from normal to a tiny planet. Some phone apps, as well as photoshop do this.</p>
<p>I think Photoshop does it by converting Cartesian coordinates to Polar coordinates.
I found a good explanation of converting them <a href="http://www.teacherschoice.com.au/maths_library/coordinates/polar_-_rectangular_conversion.htm" rel="nofollow">here</a>.</p>
<p>However, I am yet to find somewhere that describes an algorithm of the process.</p>
<p>Is there a set formula? or is it more complex than that?</p>
<p>I'm not sure if it's as easy as moving pixels and converting coordinates because I think some form of stretching must be included... hence, looking for an algorithm and this exchange as a resource. Thanks.</p>
| E.O. | 18,873 | <p>The process of creating the images you describe uses a kind of projection called <strong>stereographic projection</strong>. When you take your panorama photo, it "stores" the data on a sphere so to speak.</p>
<p>$\hspace{45mm}$<img src="https://i.stack.imgur.com/9Y3du.png" alt="enter image description here"></p>
<p>Our goal is to project that image from the sphere to the plane.</p>
<p>$\hspace{25mm}$<img src="https://i.stack.imgur.com/S5S8o.png" alt="enter image description here"></p>
<p>As you can see in the image above, the projection of a point is found by taking a line from the north pole through the point you wish to project and finding its intersection with the $z=-1$ plane. Mathematically, this translates to finding a $t$ such that
$$\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}+t\left(\begin{pmatrix}x \\ y \\ z\end{pmatrix}-\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}\right)=\begin{pmatrix}X \\ Y \\ -1\end{pmatrix},t\in\mathbb{R}$$
Here $(x,y,z)$ is the point we wish to project and $(X,Y)$ is the resulting projection. Now, if we look only at the $z$ coordinate, this boils down to solving the following equation
$$1+t(z-1)=-1$$
which has the nice and simple soltuion
$$t=\frac{2}{1-z}$$
Plugging back into our first equation we get
$$\begin{pmatrix}\frac{2x}{1-z} \\ \frac{2y}{1-z} \\ -1\end{pmatrix}=\begin{pmatrix}X \\ Y \\ -1\end{pmatrix}$$
Therefore, the coordinates for the projection is
$$(X,Y)=\left(\frac{2x}{1-z},\frac{2y}{1-z}\right)$$
Now, I don't know how the camera's actually store the data for the panorama, but assuming you have access to such an image you can transform that data onto the sphere and apply the projection to get your desired result.</p>
<p><a href="http://www.droidsage.com/wp-content/uploads/2012/11/Tiny-Planet.jpg" rel="noreferrer">Image source</a></p>
|
1,455,098 | <p>I am reading the book Foundation of Machine Learning, and the author has many proofs for different theorems. Here is one part of a proof about Perceptron algorithm which I don't quite understand</p>
<p>Note: $w_t$ is a vector</p>
<p>Let
\begin{equation}
w_{t+1} = \begin{cases}
w_t + ny_t x_t & \text{if } y_t(w.x_t) < 0 \\
w_t & \text{if } y_t(w.x_t) > 0 \\
w_t & \text{otherwise}
\end{cases}
\end{equation}</p>
<p>and in the proof, he has
\begin{equation}
\|w_{T+1}\| = \sqrt{\sum_{t \in T} \|w_{t+1}\|^2 - \| w_t \|^2}
\end{equation}</p>
<p>he mentioned that above equation uses (telescoping sum, $w_0 = 0$). I have no idea why lhs is equal to rhs using telescoping sum, can you help me explain this?</p>
<p>Thanks in advance!</p>
| Sriharsha Madala | 254,941 | <p>From my understanding of perceptron, in the training scenario, for every input, you change the weights in such a way to minimize the error (typically mean squared error). The delta weight vector is perpendicular to the current weight vector.</p>
<p>Hence $\|w_{t+1} \|^2 = \|w_{t} \|^2 + \|ny_tx_t \|^2 (\text{ if } y_t(wx_t)<0)$.</p>
<p>Writing it in the following way helps you see this more clearly:</p>
<p>\begin{align}
\|w_{T+1} \|^2 &= \|w_{T} \|^2 + \|ny_Tx_T \|^2 (\text{ if } y_T(wx_T)<0) \\
&= \|w_{T-1} \|^2 + \|ny_{T-1}x_{T-1} \|^2 (\text{ if } y_{T-1}(wx_{T-1})<0) + \|ny_Tx_T \|^2 (\text{ if } y_T(wx_T)<0) \\
&\ldots
\end{align}</p>
<p>Since $w_0$ is zero. This simplifies to the desired expression .</p>
|
1,408,231 | <p>Definition </p>
<p>A function $f$ is <strong>convex</strong> on an interval if for $a,x, \text{and} \;b$ in the interval with $a\lt x\lt b$, we have</p>
<p>$$\frac{f(x)-f(a)}{x-a}\lt \frac{f(b)-f(a)}{b-a}.$$</p>
<p>While reading the proof that if $f$ is convex on some open interval containing $a$, then $f_+'(a)$ and $f_{-}'(a)$ always exist. The proof first shows that $[f(a+h)-f(a)]/h$ is decreasing as $h\to 0^+$, so
$$f_+'(a)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}=\operatorname{inf}\{\frac{f(a+h)-f(a)}{h}:h\gt 0\}.$$</p>
<p>The proof states that this inf exists because each quotient $\frac{f(a+h)-f(a)}{h}$ for $h\gt 0$ is greater than any one such quotient for $h'\lt 0$. </p>
<p>However, I'm struggling to show this inequality, that for any $h\gt 0$ and $h'\lt 0$, we have</p>
<p>$$\frac{f(a+h')-f(a)}{h'}\lt \frac{f(a+h)-f(a)}{h}.$$</p>
<p>I would greatly appreciate it if anyone can help me prove this inequality.</p>
| zhw. | 228,045 | <p>I'm going to suggest you think geometrically here (which includes drawing a picture): Suppose $A,B,C$ are points in the plane with $A$ to the left of $B$ and $B$ to the left of $C.$ The first inequality you cite is of the form $s(A,B) \le s(A,C),$ where $s$ denotes slope. How could this be true unless $B$ lies on or below the line through $A$ and $C?$ And if $B$ satisfies that, doesn't it follow that $s(A,B) \le s(B,C)?$ The latter inquality is what you are trying to prove. So this is transparent geometrically; you "just" have to translate this picture into a bit of analysis.</p>
|
1,712,933 | <p>I am slightly ashamed to be asking this, but I have been recently reflecting on changing variables in very simple problems. If I missed a question that already discusses this please point it out to me and I will delete this one.
Anyhow writing this will probably be a learning experience.</p>
<p>Directly from the Wikipedia page on the argument I take as an example the equation:</p>
<p>$$x^6 - 9 x^3 + 8 = 0. \, $$</p>
<p>I quickly recognize this as a high school problem and use the methods that were taught to me, namely I set $x^3 = u$ so $x = u^{1/3}$.</p>
<p>Then I proceed to solve quadratic equation that results from this substitution, and only at the end I apply the reverse transformation $x^3 = u$ to get an answer for my starting variable. With not much imagination I always thought that the function used when changing variables (in the above case $f(x) = x^3)$ should be bijective in the domain of interest of the starting equation. This is because I need the inverse to return to my "starting variable".</p>
<p>But I notice on Wikipedia that a bit more is required; the change of variable function should be a diffeomorphism, we need differentiability (and even smooth manifolds for the domain and the image).</p>
<p>This is where I realized that I was never taught a proof of why the change of variables method work or how it works but I was just applying these substitutions blindly.</p>
<p>So could someone kindly point me to a source where I can improve my understanding on this very powerful method by adding rigour to what I am doing and possibly even a geometric interpretation.</p>
| Ethan Bolker | 72,858 | <p>Good question, good answers.</p>
<p>As @EricS points out, you don't have to substitute in this particular case - you can do all the work with the original variable. But you can substitute. The advantage is that changing the name of the variable in a systematic way makes the shape of the problem and solution a little clearer.</p>
<p>Since all you're doing here is finding isolated solutions, all you need is bijectivity to make sure your translation from one name space to another and back is faithful.</p>
<p>If you want to do more than algebra you may need a better dictionary - that is, a substitution with better properties. If as @mweiss comments you want to do calculus on the transformed equation then the substitution and its inverse must be differentiable. That's essentially what the wikipedia page is saying, in a more abstract context.</p>
<p>When you study abstract algebra you'll want your "substitutions" to respect the algebraic properties of the domain and range. That's the essence of @asymplectomorphic 's comment about linear algebra.</p>
|
95,343 | <p>I am writing a code for a Lagrange multipliers problem with three variables.
My code looks like this:</p>
<pre><code> f[x_, y_, z_] := (m1 x + m2 y + m3 z)/Sqrt[m1^2 x + m2^2 y + m3^2 z]
g[x_, y_, z_] := x + y + z - 1
Minimize[{f[x, y, z], g[x, y, z] == 0,
m1 > 0, m2 > 0, m3 > 0, x > 0, y > 0, z > 0}, {x, y, z}]
</code></pre>
<p>But it keeps running with no answer, why is that?</p>
| Community | -1 | <p>To confirm @belisarius comment:</p>
<pre><code>f[x_, y_, z_] := (m1 x + m2 y + m3 z)/Sqrt[m1^2 x + m2^2 y + m3^2 z]
g[x_, y_, z_] := x + y + z - 1
L = f[x, y, z] + λ g[x, y, z];
Solve[{Grad[L, {x, y, z}] == 0, g[x, y, z] == 0, m1 > 0, m2 > 0,
m3 > 0, x > 0, y > 0, z > 0}, {x, y, z, λ}]
{}
Simplify[Reduce[{Grad[L, {x, y, z}] == 0, g[x, y, z] == 0, m1 > 0,
m2 > 0, m3 > 0, x > 0, y > 0, z > 0}, {x, y, z, λ}],
m1 > 0 && m2 > 0 && m3 > 0]
</code></pre>
<p><a href="https://i.stack.imgur.com/cXkop.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cXkop.png" alt="enter image description here"></a></p>
<p>calculated with Mma 10.2 on Windows 10</p>
|
3,439,833 | <p><strong>Problem:</strong>
Let <span class="math-container">$f:\mathbb{R} \rightarrow \mathbb{R}$</span> be differentiable.
Suppose <span class="math-container">$f(0)=0$</span>, and <span class="math-container">$f'(x) < \frac{1}{2}$</span> for all <span class="math-container">$a$</span>. Show <span class="math-container">$f(4) < 2$</span>.</p>
<p>This question is intuitively pretty simple, but I'm not sure what is generally regarded as sufficiently rigorous.</p>
<p>Simply, if the slope of the function never reaches <span class="math-container">$\frac{1}{2}$</span>, then combining all these small intervals will produce <span class="math-container">$f(4) <2$</span>.</p>
<p><span class="math-container">$\forall a \in \mathbb{R}, \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a} < \frac{1}{2} \implies \forall a \in \mathbb{R}, f(x)-f(a) < \frac{1}{2}(x-a)$</span> </p>
<p>Letting <span class="math-container">$a=0, x=4$</span>, we have <span class="math-container">$f(4) < 2$</span>.</p>
<p><strong>My Question:</strong></p>
<p>I'm not sure if I'm skipping a ton of steps in that implication above. It looks like I just jumped to the conclusion without proving anything.</p>
<p>Is what I'm missing some sort of "interval argument"?</p>
<p>For example:
If I translated using the delta-epsilon definition of limits
<span class="math-container">$$\forall a \in \mathbb{R}, \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$$</span></p>
<p>to <span class="math-container">$f(x)-f(a) < \frac{1}{2}(x-a)$</span> for some open interval <span class="math-container">$ x \in (a- \delta, a+ \delta)$</span>.</p>
<p>I would still need to show that the union of <em>infinitely</em> many of such small intervals satisfies <span class="math-container">$f(x)-f(a) < \frac{1}{2}(x-a)$</span>. How do I write this rigorously?</p>
| Peter Szilas | 408,605 | <p><span class="math-container">$f(4)-f(0)=\displaystyle{\int_{0}^{4}}f'(x)dx< $</span></p>
<p><span class="math-container">$\displaystyle{\int_{0}^{4}}(1/2)dx =(1/2)4=2$</span>.</p>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Mark Wildon | 7,709 | <p>Let <span class="math-container">$f : \mathbb{F}_2^n \rightarrow \mathbb{F}_2$</span> be a boolean function. Ordering <span class="math-container">$\mathbb{F}_2$</span> so that <span class="math-container">$0<1$</span>, we say that <span class="math-container">$f$</span> is <em>monotone</em> if <span class="math-container">$f(x_1,\ldots,x_n) \le f(y_1, \ldots, y_n)$</span> whenever <span class="math-container">$x_1 \le y_1, \ldots, x_n \le y_n$</span>. The <em>Dedekind number</em> <span class="math-container">$M(n)$</span> is the number of monotone Boolean functions of <span class="math-container">$n$</span> variables, up to logical equivalence.</p>
<p>Equivalently, <span class="math-container">$M(n)$</span> is the number of Boolean functions (up to logical equivalence) that can be expressed without negation, or the number of <a href="https://en.wikipedia.org/wiki/Access_structure" rel="noreferrer">access structures</a> on <span class="math-container">$n$</span> people. For example, </p>
<p><span class="math-container">$$(x_1 \wedge x_2) \vee (x_1 \wedge x_3) \vee (x_2 \wedge x_3)$$</span> </p>
<p>is counted by <span class="math-container">$M(3)$</span>, and corresponds to the access structure where any pair of people, or all three people, form an authorized set. In general the minimal authorized sets form an antichain in the poset of subsets of <span class="math-container">$\{1,\ldots, n\}$</span>, so <span class="math-container">$M(n)$</span> is the number the number of antichains in <span class="math-container">$\mathcal{P}(\{1,\ldots,n\})$</span>. Since the maximal subsets in an <a href="https://en.wikipedia.org/wiki/Abstract_simplicial_complex" rel="noreferrer">abstract simplicial complex</a> form an antichain, <span class="math-container">$M(n)$</span> is also the number of abstract simplicial complexes on <span class="math-container">$\{1,\ldots, n\}$</span>. </p>
<p>The antichain interpretation makes it plausible that</p>
<p><span class="math-container">$$\log M(n) \sim \binom{n}{\frac{n}{2}} \sim \sqrt{\frac{2}{\pi}} \frac{2^n}{\sqrt{n}} $$</span></p>
<p>and this was <a href="http://www.ams.org/journals/proc/1969-021-03/S0002-9939-1969-0241334-6/S0002-9939-1969-0241334-6.pdf" rel="noreferrer">proved by Kleitman</a>. </p>
<p>An important open problem (mentioned in the comments above) is to give a more precise asymptotic estimate for <span class="math-container">$M(n)$</span>, and to determine the value of <span class="math-container">$M(n)$</span> for small <span class="math-container">$n$</span>. At the moment, <span class="math-container">$M(n)$</span> is <a href="https://arxiv.org/pdf/1209.4623.pdf" rel="noreferrer">known precisely</a> only for <span class="math-container">$n \le 7$</span>. The linked paper has the best known asymptotic formula, due to Korshunov.</p>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Mike Earnest | 59,232 | <p>A ranked poset is called a <em>symmetric chain order</em>, or SCO, if it can partitioned into rank symmetric, saturated chains. Such posets are particularly nice, as they are rank-symmetric, unimodal, and have the strong Sperner property. If <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> are SCO's, then so is <span class="math-container">$P\times Q$</span>. In particular, the Boolean poset <span class="math-container">${\bf 2}^n$</span> of subsets of <span class="math-container">$\{1,2,\dots,n\}$</span>, which is the product of <span class="math-container">$n$</span> copies of a two-element chain, is an SCO.</p>
<p>Let <span class="math-container">$G$</span> be a subgroup embedded in the symmetric group <span class="math-container">$S_n$</span>. This induces an action on <span class="math-container">${\bf 2}^n$</span> where <span class="math-container">$gS=\{g(s):s\in S\}$</span>. Define the quotient poset <span class="math-container">${\bf 2}^n/G$</span> on the orbits of this action, with <span class="math-container">$[S]\le [T]$</span> if some <span class="math-container">$S'\subseteq T'$</span> for <span class="math-container">$S'\in[S]$</span> and <span class="math-container">$T'\in [T]$</span>. </p>
<p>Several familiar posets are realized as quotients of <span class="math-container">${\bf 2}^n$</span>.</p>
<ul>
<li><p><span class="math-container">$L(m,k)$</span>, the sub-poset of Young's lattice consisting of partitions which fit in an <span class="math-container">$m\times k$</span> box. Here, <span class="math-container">$n=mk$</span>, and the subgroup is the wreath product <span class="math-container">$S_m\wr S_k$</span>.</p></li>
<li><p>the poset of isomorphism classes finite simple graphs on <span class="math-container">$n$</span> vertices with the subgraph relation.</p></li>
</ul>
<blockquote>
<p><strong>Conjecture:</strong> For all <span class="math-container">$n\ge 0$</span> and all <span class="math-container">$G\le S_n$</span>, <span class="math-container">${\bf 2}^n/G$</span> is an SCO.</p>
</blockquote>
<p>Partial results:</p>
<ul>
<li><p><span class="math-container">$L(4,n)$</span> is an SCO.</p></li>
<li><p>Letting <span class="math-container">$G=\langle (1\,2\,\dots\,n)\rangle$</span>, then the necklace poset <span class="math-container">${\bf 2}^n/G$</span> is an SCO.</p></li>
</ul>
|
322,598 | <p><a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">Partially ordered sets</a> (<em>posets</em>) are important objects in combinatorics (with <a href="https://gilkalai.wordpress.com/2019/02/05/extremal-combinatorics-v-posets/" rel="noreferrer">basic connections to extremal combinatorics</a> and to algebraic combinatorics) and also in other areas of mathematics. They are also related to <em>sorting</em> and to other questions in the theory of computing. I am asking for a list of open questions and conjectures about posets.</p>
| Joseph Van Name | 22,277 | <p>So in mathematics, one can iterate operation of mapping a group to its automorphism group transfinitely to obtain the automorphism tower of a group. And <a href="https://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1336588247" rel="nofollow noreferrer">the automorphism tower always terminates</a>. The next natural follow up question to this result is to ask about the congruence lattice tower where one begins with a lattice.</p>
<p>Now, if <span class="math-container">$L$</span> is a lattice, then the mapping <span class="math-container">$e:L\rightarrow\mathrm{Con}(L)$</span> defined by letting <span class="math-container">$(x,y)\in e(a)$</span> if and only if
<span class="math-container">$x\vee a=y\vee a$</span> is a lattice homomorphism precisely when the lattice <span class="math-container">$L$</span> is distributive. Since we want <span class="math-container">$L$</span> to embed in <span class="math-container">$\mathrm{Con}(L)$</span>, we want to restrict
our attention to distributive lattices. Furthermore, if <span class="math-container">$L$</span> is a lattice, then <span class="math-container">$\mathrm{Con}(L)$</span> is always a distributive lattice, so we have no choice but to only
consider distributive lattices in the congruence lattice tower problem. It is a not too difficult exercise to show that the mapping <span class="math-container">$e:L\rightarrow\mathrm{Con}(L)$</span>
is a lattice isomorphism if and only if <span class="math-container">$L$</span> is a finite Boolean algebra. Therefore, the congruence tower problem is quite easy and boring for the variety of all
distributive lattices. The congruence tower problem for frames instead of distributive lattices is a difficult open question of mathematical significance.</p>
<p>A frame is a complete lattice <span class="math-container">$L$</span> that satisfies the infinite distributivity law
<span class="math-container">$x\wedge\bigvee R=\bigvee_{r\in R}(x\wedge r)$</span>. The frames <span class="math-container">$L$</span> are precisely the complete lattices which happen to be complete Heyting algebras. If <span class="math-container">$L$</span> is a frame, then an equivalence relation <span class="math-container">$\simeq$</span> on <span class="math-container">$L$</span> is said to be a frame congruence if <span class="math-container">$\bigvee_{i\in I}x_{i}\simeq\bigvee_{i\in I}y_{i}$</span> whenever <span class="math-container">$x_{i}\simeq y_{i}$</span> for <span class="math-container">$i\in I$</span> and <span class="math-container">$r\wedge s\simeq t\wedge u$</span> whenever <span class="math-container">$r\simeq t,s\simeq u$</span>. </p>
<p>If <span class="math-container">$L$</span> is a frame, then let <span class="math-container">$\mathfrak{C}(L)$</span> denote the collection of all congruences of the frame <span class="math-container">$L$</span>. Then <span class="math-container">$\mathfrak{C}(L)$</span> is itself a frame, and there is a mapping <span class="math-container">$e:L\rightarrow\mathfrak{C}(L)$</span> where <span class="math-container">$(x,y)\in e(a)$</span> if and only if <span class="math-container">$x\vee a=y\vee a$</span> which happens to be a frame homomorphism. </p>
<p>One can define <span class="math-container">$\mathfrak{C}^{\alpha}(L)$</span> for all ordinals <span class="math-container">$\alpha$</span> by letting
<span class="math-container">$\mathfrak{C}^{\alpha+1}(L)=\mathfrak{C}(\mathfrak{C}^{\alpha}(L))$</span> for all <span class="math-container">$\alpha$</span> and where for limit ordinals <span class="math-container">$\gamma$</span>, we have <span class="math-container">$\mathfrak{C}^{\gamma}(L)=\varinjlim_{\alpha<\gamma}\mathfrak{C}^{\alpha}(L)$</span> (direct limits are taken in the category of frames).</p>
<p>If <span class="math-container">$e:L\rightarrow\mathfrak{C}^{\alpha}(L)$</span> is the canonical mapping, then for each complete Boolean algebra <span class="math-container">$B$</span>, the mapping <span class="math-container">$\mathrm{Hom}(\mathfrak{C}^{\alpha}(L),B)\rightarrow\mathrm{Hom}(L,B)$</span> in the category of frames where <span class="math-container">$\phi\mapsto\phi\circ e$</span> is a bijection (this result should convince you that <span class="math-container">$\mathfrak{C}^{\alpha}(L)$</span> is important).</p>
<p>So there are frames <span class="math-container">$L$</span> where the tower <span class="math-container">$(\mathfrak{C}^{\alpha}(L))_{\alpha}$</span> never stops growing. If <span class="math-container">$B$</span> is a complete Boolean algebra, then the canonical mapping <span class="math-container">$e:B\rightarrow\mathfrak{C}(B)$</span> is an isomorphism, and if the tower <span class="math-container">$(\mathfrak{C}^{\alpha}(L))_{\alpha}$</span> stops growing, then <span class="math-container">$\mathfrak{C}^{\alpha}(L)$</span> is a complete Boolean algebra for large enough <span class="math-container">$\alpha$</span>.</p>
<p>Now, in all known examples, either <span class="math-container">$(\mathfrak{C}^{\alpha}(L))_{\alpha}$</span> never stops growing or the tower <span class="math-container">$(\mathfrak{C}^{\alpha}(L))_{\alpha}$</span> stops growing at or before around the 4th step. Suppose that <span class="math-container">$\alpha$</span> is an ordinal. Then is there a frame <span class="math-container">$L$</span> where <span class="math-container">$\alpha$</span> is the least ordinal with <span class="math-container">$\mathfrak{C}^{\alpha}(L)=\mathfrak{C}^{\beta}(L)$</span> for <span class="math-container">$\beta\geq\alpha$</span> in the sense that the canonical frame homomorphism between these frames is an isomorphism?</p>
|
244,875 | <p>\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Note that here $x$ is unrestricted. I need to prove that the dual of this program is given by
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA\leq c^T
\end{align}</p>
<p>But in the constraint, I always get an equality (using what I learnt)
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA = c^T
\end{align}
Please give some explanation also. </p>
| Keith Reyes | 49,970 | <p>Given </p>
<p>$\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}$</p>
<p>I think the dual of the problem would be</p>
<p>$\begin{align}
\max_{x} b^Tw \\
s.t.~A^Tw (*)c
\end{align}$</p>
<p>where $(*)$ actually depends on the restrictions of the variables in the primal problem. If the restriction is $x_i \geq 0$ then the type of constraint will follow the same inequality. But if $x_i$'s are unrestricted, then the constraints will become $=$</p>
|
2,509,583 | <p>How is intermediate value theorem valid for $\sin x$ in $[0,\pi]$?</p>
<p>It has max value $1$ in the interval $[0,\pi]$ which doesn't lie between values given by $\sin0$ and $\sin\pi$.</p>
| gt6989b | 16,192 | <p><a href="https://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow noreferrer">Intermediate Value Theorem</a> says that </p>
<blockquote>
<p>Let $f$ be a function, continuous on the interval, $[a, b]$. Then $f$ takes any value between $f(a)$ and $f(b)$ at some point within the interval.</p>
</blockquote>
<p>In your case, $\sin x$ is 0 at both $x=0$ and $x=\pi$, and the IVT implies there is a point $y\in[a,b]$ such that $\sin y = 0$. You can let $y=0$ or $y=\pi$...</p>
<p><strong>UPDATE</strong></p>
<p>Note this does <em>not</em> mean $f$ must always be between $f(a)$ and $f(b)$, just that every value between $f(a)$ and $f(b)$ is covered.</p>
<hr>
|
3,603,926 | <p>I came across this elementary counting problem: A coffee shop has 4 different types of coffee. You can order your coffee in a small, medium, or large cup. You can also choose whether you want to add cream, sugar, or milk (any combination is possible, for example, you can choose to add all three). In how many ways can you order your coffee?</p>
<p>My attempt:
<span class="math-container">$4*3*2^3$</span>. Is that correct? </p>
| Arthur | 15,500 | <p>Yup, that is the answer. You have four options for coffee, three options for size, two options for cream, two options for sugar, and two options for milk, and you have to pick one option for each of these five choices. This does indeed give <span class="math-container">$$4\cdot 3\cdot 2\cdot 2\cdot 2$$</span></p>
|
221,137 | <p>What's the difference between Fourier transformations and Fourier Series? </p>
<p>Are they the same, where a transformation is just used when its applied (i.e. not used in pure mathematics)?</p>
| Shivam Kumar | 90,692 | <p>Fourier transform is used to transform periodic and non-periodic signals from time domain to frequency domain. It can also transform Fourier series into the frequency domain, as Fourier series is nothing but a simplified form of time domain periodic function.</p>
<h3>Fourier series</h3>
<ol>
<li><p>Periodic function => converts into a <em>discrete</em> exponential or sine and cosine function. </p></li>
<li><p>Non-periodic function => not applicable</p></li>
</ol>
<h3>Fourier transform</h3>
<ol>
<li><p>Periodic function => converts its Fourier series in the frequency domain. </p></li>
<li><p>non-Periodic function => converts it into <em>continuous</em> frequency domain.</p></li>
</ol>
|
2,948,673 | <p>I was thinking about the squeeze theorem here. We can denote the <span class="math-container">$\left[\frac{1}{|x|}\right] =n$</span>, and then try something like:</p>
<p><span class="math-container">$$x^2(1+2+3+...+(n-1)+(n-1)) \leq x^2 \frac{n(n+1)}{2} \leq x^2(1+2+3+...+(n-1)+(n+1))$$</span></p>
<p>But I don't know what to do with <span class="math-container">$x^2$</span>. Given that <span class="math-container">$\left[ \frac{1}{|x|} \right]=n$</span>, how do we proceed to find <span class="math-container">$x^2$</span>?</p>
| MPW | 113,214 | <p>Why not just substitute <span class="math-container">$x=1/n$</span> (noting that the limits from both sides are the same) and consider <span class="math-container">$$\lim\limits_{n\to\infty} \frac1{n^2}(1+2+\cdots+n)$$</span>
instead?</p>
|
1,221,368 | <p>An inheritance is divided between three people in the ratio $4:7:2$. If the least amount received is $\$\ 2300$ calculate how much the other two people received?</p>
| direprobs | 226,619 | <p>The answer is:</p>
<p>we have 13 parts in the total amount: 4 + 7 + 2 = 13 parts
The lowest person receives 2 parts that is 2300 and so 2300/2= 1150 for each part. Thus the total amount is (1150 * 13) = 14950</p>
<p>Therefore:</p>
<blockquote>
<p>2/13 * 14950 = 2300</p>
<p>4/13 * 14950 = 4600</p>
<p>7/13 * 14950 = 8050</p>
<p>2300 + 4600 + 8050 = 14950</p>
</blockquote>
|
103,358 | <p>I am trying to illustrate some simple ideas with exponents. I can manually express something like $5^4$ as $5 \cdot 5 \cdot 5 \cdot 5$, but wondered how to get <em>Mathematica</em> to do that for me.</p>
<p>I found the example below in the documentation, but can't figure out how to "massage" it to work with a number that only has one factor, for example I would like $625$ to be represented as $5 \cdot 5 \cdot 5 \cdot 5$:</p>
<pre><code>CenterDot @@ (Superscript @@@ FactorInteger[20!])
</code></pre>
<p>Any ideas would be appreciated.</p>
| SC_L | 30,459 | <p>I got the same problem and solved it by doing this. Just go to the <strong>Preferences</strong>-><strong>Internet Connectivity</strong> and turn off <strong>Allow the Wolfram System to access the Internet</strong>, then relaunch Mathematica.</p>
|
103,358 | <p>I am trying to illustrate some simple ideas with exponents. I can manually express something like $5^4$ as $5 \cdot 5 \cdot 5 \cdot 5$, but wondered how to get <em>Mathematica</em> to do that for me.</p>
<p>I found the example below in the documentation, but can't figure out how to "massage" it to work with a number that only has one factor, for example I would like $625$ to be represented as $5 \cdot 5 \cdot 5 \cdot 5$:</p>
<pre><code>CenterDot @@ (Superscript @@@ FactorInteger[20!])
</code></pre>
<p>Any ideas would be appreciated.</p>
| level1807 | 5,784 | <p>This annoying bug has finally been fixed in 11.0</p>
|
359,816 | <p>Let $(f_n)$ and $(g_n)$ be sequences of nonnegative function in $L^1(\mathbb R)$, for which
$$
f_n \to 0, \\
g_n \to 0,
$$
almost everywhere.
Show
$$
\int_A \frac{2f_n g_n}{1+f_n^2+g_n^2} \to 0,
$$
when $A \subset \mathbb R$ is a set of finite measure.</p>
<p>Define
$$
h_n = \frac{2f_n g_n}{1+f_n^2+g_n^2}.
$$
I know
$$
h_n \le 2f_n g_n,
$$
but I can't use a convergence theorem, because the product of integrable function may fail to be integrable.<br>
I considered uniform integrability
$$
\text{Since }
f_n \in L^1: \forall \epsilon, \exists \delta : m(B) < \delta \to \int_B f_n <\epsilon,
$$
where $m(\cdot)$ is Lebesgue measure.<br>
To use that, I write $A=\cup_m B_{n,m}$ where the index $n$ is associated to $f_n$.<br>
I can't go further.</p>
| Umberto P. | 67,536 | <p>You have $0 \le h_n \le g_n$ since $\dfrac{2f_n}{1 + f_n^2} \le 1$. Thus $h_n \to 0$. You also have $0 \le h_n \le 1$ since $\dfrac{2f_n g_n}{1 + f_n^2 + g_n^2} \le 1$. Since $A$ has finite measure, this provides a majorant of the sequence.</p>
|
121,760 | <p>I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.</p>
<p>With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.</p>
<p>But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find the minimal polynomial ? </p>
| Zarrax | 3,035 | <p>Let $\alpha = \sqrt[3]{7 - \sqrt{2}}$ and $P(x) = x^6 - 14x^3 + 47$. Since $P(\alpha) = 0$, in order to show $P$ is irreducible it suffices to show $[Q(\alpha):Q] = 6$. Since $7 - \alpha^3 = \sqrt{2}$, you have that $\sqrt{2} \in Q(\alpha)$ and thus
$$[Q(\alpha):Q] =[Q(\alpha):Q(\sqrt{2})][Q(\sqrt{2}):Q] = 2[Q(\alpha):Q(\sqrt{2})]$$
Thus our goal is to show that $[Q(\alpha):Q(\sqrt{2})] = 3$. Since $\alpha$ satisfies the third-degree polynomial equation $Q(x) = x^3 - 7 + \sqrt{2} = 0$ over $Q(\sqrt{2})$, our task is equivalent to showing $Q(x)$ is irreducible over $Q(\sqrt{2})$.</p>
<p>But if $Q(x)$ were not irreducible over $Q(\sqrt{2})$, it would have a linear factor. Thus it would have a root $a + b\sqrt{2}$, with $a$ and $b$ rational. But the roots of $Q(x) = 0$ are the numbers $\alpha$, $e^{2\pi i \over 3}\alpha$, and $e^{4\pi i \over 3}\alpha$. Of these, only $\alpha$ is real, so we have
$$\alpha = a + b\sqrt{2}$$
Cubing, we get
$$7 - \sqrt{2} = (a + b\sqrt{2})^3$$
Thus we must have
$$7 + \sqrt{2} = (a - b\sqrt{2})^3$$
Multiplying the above two equations together we get
$$47 = (a^2 - 2b^2)^3$$
Since $47$ is not the cube of a rational number, we have arrived at a contradiction and we are done.</p>
|
1,425,961 | <p>I am about to take a course in undergraduate Algebra which is using Artin's Algebra. I have been studying from it and haven't really appreciated the combination of text and problems (granted, I haven't gone too deep into the text). </p>
<p>Now, this is the text for the class, so I would expect a nice compliment from the lecture to go along with Artin. I was thinking about getting Dummit and Foote, which seems to have high praise across the board. What are the strengths and weaknesses of Dummit and Foote, compared to Artin? Is there a better text for a class at this level?</p>
<p>EDIT: I really don't want a recommendation for another text, but how does Artin compare to other texts. </p>
| Community | -1 | <p>I have and am still frequently using both books and I have to say the main thing they have going is that :</p>
<ul>
<li>D&F has great exercises.</li>
<li>Artin is friendlier to the audience.</li>
</ul>
<p>My usual advice is that one learns from Artin and then does the exercises in D&F, but overall they are both great books. From what people have said to me, D&F is better as a book to have after you already understand the content, for review it is great, but on a first pass it may not be so good. I have heard similar things in that last regard, about the textbook Rudin's principles of mathematical analysis (just for comparison if you have read it).</p>
|
483,602 | <p>$$
\large\sqrt{8+\sqrt{5}-\sqrt{6-2\sqrt{5}}}
$$</p>
<p>I am asked to write the following as a ratio of two integers (i.e.,) show that this value is rational
(Do not use calculator)</p>
<p>What I tried doing was putting everything to the exponent of $.5$ and then was unsure how to proceed </p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> $6-2\sqrt{5}=(\sqrt{5}-1)^2$. </p>
|
1,808,019 | <p>I'm trying to better understand how the following group action looks like on a torus. The group $\mathbb{R}$ acts on the torus $C\times C=\{(e^{ix},e^{iy}):x,y\in \mathbb{R}\}$ ($C$ being the unit circle in $\mathbb{C}$) via $$t\cdot (e^{ix},e^{iy})\mapsto (e^{i(x+t)},e^{iy})$$</p>
<p>I am able (I think) to visualize the torus based on the definition of $C\times C$: the first coordinate gives us the ring of unit length in the donut and, as $x$ sweeps through $\mathbb{R}$, the second coordinate gives us the "hollow body" of the donut, again of unit length, as $y$ sweeps through $\mathbb{R}$. So here is where the problems begin for me.</p>
<ol>
<li><p>What am I doing wrong to show that this a homomorphism? $$st\cdot (e^{ix},e^{iy})\mapsto (e^{i(x+st)},e^{iy})$$ but $$[s\cdot (e^{ix},e^{iy})][t\cdot (e^{ix},e^{iy})] = (e^{i(x+s)},e^{iy})(e^{i(x+t)},e^{iy}) = (e^{i(2x+s+t)},e^{i2y})$$</p></li>
<li><p>For $t\in \mathbb{R}$, the orbit $\mathbb{R}(t)$ is $\{(e^{i(x+t)},e^{iy}):t\in \mathbb{R}\}$. Is this any different from $C\times C$?</p></li>
</ol>
| Surb | 154,545 | <p><strong>Just a complement : Proof using combinatorial argument</strong></p>
<p>Let $X$ a set with $x$ elements, $Y$ set with $y$ elements s.t. $X\cap Y=\emptyset$ and $N$ a set with $n$ element.
$$(x+y)^n=\#\{f:N\to X\cup Y\mid f\text{ is a function}\}.$$</p>
<p>An other way to count such a function is the following one. Let $f_k$ a function that through $k$ elements in $X$ and $n-k$ elements in $y$. So, it choose $\binom{n}{k}$ element of $n$, and each such element has $x$ possibilities. Then you take the other $\binom{n-k}{n-k}$ element, and each element has $y$ possibilities. Then, $$\#\{f_k:N\longrightarrow X\cup Y\}=\binom{n}{k}x^k\binom{n-k}{n-k}y^{n-k}=\binom{n}{k}x^ky^{n-k}.$$
Finally
$$(x+y)^n=\#\{f:N\to X\cup Y\mid f\text{ is a function}\}=\sum_{k=0}^n\#\{f_k:N\longrightarrow X\cup Y\}$$
$$=\sum_{k=0}^n \binom{n}{k}x^ky^{n-k}$$</p>
|
70,158 | <p>This question is related to <a href="https://mathoverflow.net/questions/52825/coloring-mathbb-zk"> this one </a> but feels more Ramsey-type, so perhaps it is easier. Let $S$ be a finite set, $|S|=k$. Suppose we color all subsets of $S$ in $1000$ colors. What is the maximal (in terms of $k$) guaranteed length $l=l(k)$ of a monochromatic sequence of pairwise different subsets $A_1,A_2,..., A_l$ such that $|A_i\setminus A_{i+1}|+|A_{i+1}\setminus A_i|\le 2$ for every $i$? Clearly if $A$ is a subset of $S$ such that all 2-element subsets of $A$ are monochromatic, then $l(n)\ge |A|-1$ (there is a sequence of 2-element subsets of $A$ which satisfies the above property). So $l(k)$ is at least as big as the corresponding number from the Ramsey theory. Is it much bigger? The number 1000 is of course "any fixed number". </p>
<p><b> Update 1 </b> Fedor and Tony showed below that $l(k)\ge k/1000$. Thus only the first question remains: What is $l(k)$? Is it exponential in $k$, for example?</p>
<p><b> Update 2 </b> Although the question I asked makes sense (see Update 1), I realized that it is not the question I meant to ask. Here is the correct question. Same assumptions: $|S|=k$, 1000 colors. We consider monochromatic sequences of pairwise different subsets ${\mathcal A}=A_1,A_2,...,A_l$, where $|A_i\setminus A_{i+1}|+|A_{i+1}\setminus A_i|\le 2$. For each of these sequences we compute $\chi({\mathcal A})=|A_1\setminus A_l|+|A_l\setminus A_1|$. Now the question: what is the maximal guaranteed $\chi({\mathcal A})$ in terms of $k$, call it $\chi(k)$? By Ramsey, this number grows with $k$. Indeed if we color just $s$-element subsets, we will be able (if $k\gg s$) to find a subset of size $2s$ where all subsets of size $s$ are colored with the same color; then we can find a monochromatic sequence of subsets of size $s$ with the above property and $\chi=2s$ because the first and the last subsets in that sequence are disjoint. The question is what is the growth rate of $\chi(k)$. The question is motivated by Justin Moore's answer
<a href="https://mathoverflow.net/questions/37449/covers-of-z-infty"> here.</a> </p>
| Tony Huynh | 2,233 | <p>Here is a proof of a very weak upper bound for $l(k)$. Consider the colouring of $2^S$ where each set is coloured by its size (mod 1000). A good monochomatic sequence must consist of sets of the same size. Thus we obtain $l(k) \leq \binom{k}{k/2}$. </p>
<p>However, we can be a bit smarter. Instead of colouring <em>all</em> $i$-subsets of $S$ with the same colour, we can use 333 colours and still guarantee that a good monochromatic sequence must use sets of the same size. Thus, we are lead to the problem of 333-colouring $\binom{S}{i}$ to minimize the length of a good monochromatic sequence inside $\binom{S}{i}$. </p>
|
3,669,961 | <p>Let <span class="math-container">$f:[a,b] \longrightarrow \mathbb{R}$</span> be a integrable function. I know if <span class="math-container">$ f> 0 $</span> then
<span class="math-container">$$\int_{a}^{b} f(x)\; dx >0.$$</span> The converse is true? That is, if <span class="math-container">$$\int_{a}^{b} f(x)\; dx >0$$</span> then <span class="math-container">$f>0$</span>?
I couldn't think of an example that makes it false the converse.</p>
| José Carlos Santos | 446,262 | <p>What about <span class="math-container">$\displaystyle\int_{-1}^3x\,\mathrm dx$</span>, which happens to be equal to <span class="math-container">$4$</span>?</p>
|
389,912 | <p>I'm writing a paper and want to cite some references to efficiently prove that over any field <span class="math-container">$k$</span> of characteristic zero, every irreducible representation of a product of symmetric groups, say</p>
<p><span class="math-container">$$ S_{n_1} \times \cdots \times S_{n_p} $$</span></p>
<p>is isomorphic to a tensor product <span class="math-container">$\rho_1 \otimes \cdots \otimes \rho_p$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$S_{n_i}$</span>.</p>
<p>I have an open mind about this, but I'm imagining doing it by finding references for these two claims:</p>
<ol>
<li><p>If <span class="math-container">$k$</span> is an algebraically closed field of characteristic zero, every irreducible representation of a product <span class="math-container">$G_1 \times G_2$</span> of finite groups is of the form <span class="math-container">$\rho_1 \otimes \rho_2$</span> where <span class="math-container">$\rho_i$</span> is an irreducible representation of <span class="math-container">$G_i$</span>.</p>
</li>
<li><p>If <span class="math-container">$k$</span> has characteristic zero and <span class="math-container">$\overline{k}$</span> is its algebraic closure, every finite-dimensional representation of <span class="math-container">$S_n$</span> over <span class="math-container">$\overline{k}$</span> is isomorphic to one of the form <span class="math-container">$\overline{k} \otimes_k \rho$</span> where <span class="math-container">$\rho$</span> is a representation of <span class="math-container">$S_n$</span> over <span class="math-container">$k$</span>.</p>
</li>
</ol>
<p>Serre's book <em>Linear Representations of Finite Groups</em> states the first fact for <span class="math-container">$k = \mathbb{C}$</span> but apparently not for a general algebraically closed field of characteristic zero. (It's Theorem 10.) It could be true already for any field of characteristic zero, which would simplify my life.</p>
<p>The second fact should be equivalent to saying that <span class="math-container">$\mathbb{Q}$</span> is a splitting field for any symmetric group, which seems to be something everyone knows - yet I haven't found a good reference.</p>
| Community | -1 | <p>For number 1, when a statement in the representation theory of finite groups is true over <span class="math-container">$\mathbb{C}$</span>, I think it should be true for any algebraically closed field of characteristic zero. That's because the development of the representation theory of finite groups proceeds the same way over any algebraically closed field of characteristic zero. I don't know a precise statement of this, but in algebraic geometry, a similar idea is called the "Lefschetz principle".</p>
|
191,262 | <p>I have to fit a Cos^2 function to data I measured. The function is <span class="math-container">$a \cos^2(\frac{bx\pi}{180}+c)+d $</span> and I tried the FindFit and Linear Model Function. I have 4 datasets which I have to fit, the first one worked. The other three only yielded not usable fits. I am pretty new to Mathematica so I hope its just a newbie mistake which is easy to fix.</p>
<p>Here's a minimal example: </p>
<pre><code>data45 = Import["data45.txt", "table"]
{{0, 132}, {20, 279.5}, {40, 289}, {60, 312}, {80, 307}, {100,
173}, {120, 92}, {140, 25}, {160, 44.5}, {180, 109.5}, {200,
230.5}, {220, 305}, {240, 339}, {260, 246.5}, {280, 181.5}, {300,
92.5}, {320, 32}, {340, 43}}
FindFit[data45, a Cos[(b x Pi)/180 + c]^2 + d, {a, b, c, d}, x]
{a -> 45.2733, b -> 0.886263, c -> 39.01, d -> 157.974}
</code></pre>
<p>This yields the following fit of the data: </p>
<p><a href="https://i.stack.imgur.com/ZXm5Q.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZXm5Q.png" alt="Fit"></a></p>
<p>Which is not usable. </p>
<p>I would really appreciate some help! </p>
<p>Greetings</p>
| JimB | 19,758 | <p>The two answers so far (@MassDefect and @m0nhawk) are exactly what you want to do: have good initial guesses (especially for anything dealing with sines and cosines).</p>
<p>If you are fitting many sets of data to the same model, then automating the initial values is recommended. For your example, you could use:</p>
<pre><code>sol = FindFit[data45, a Cos[(b x Pi)/180 + c]^2 + d,
{{a, (Max[data45[[All, 2]]] - Min[data45[[All, 2]]])/2}, b, c, {d, Mean[data45[[All, 2]]]}}, x]
</code></pre>
|
1,005,921 | <p>I was just wondering if anyone could shed more light on specific topics in applied mathematics or other skills (programming, etc) commonly used in the following industries: oil and gas, aviation/aeronautics. In particular, I'm interested in how mathematical modelling could be applied in these industries (from the perspective of a Maths PhD student). Would be quite interested to hear from others with experience in any of these industries. I've heard a lot about CFD in aeronautics-I was wondering if there were any others. </p>
<p>Thanks</p>
| 5xum | 112,884 | <p>I don't know about digits of $\pi$, but if you have a truly random generator, then "lags" of certain digits will not correct themselves.</p>
<p>By that I mean the following:</p>
<p>Say you have a random generator that returns integers from $0$ to $9$, and say that in the last one million tries, the generator generated $200.000$ nines. Then the probability of the generator generating another nine in the next step is still $\frac1{10}$, meaning that the probability that there will be less nines than other numbers in the <em>next</em> million integers generated is exactly the same as it would be if the generator produced zero nines in the first million tries.</p>
|
2,744,172 | <p>(Note: This question is tangentially related to this <a href="https://math.stackexchange.com/q/3609304">later one</a>.)</p>
<p>Let <span class="math-container">$$\sigma(x) = \sum_{d \mid x}{d}$$</span> denote the <em>sum of divisors</em> of <span class="math-container">$x \in \mathbb{N}$</span>, where <span class="math-container">$\mathbb{N}$</span> is the set of natural numbers or positive integers.</p>
<p>Recall that a <em>Descartes number</em> is an odd number <span class="math-container">$n = km$</span>, with <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>, satisfying <span class="math-container">$$\sigma(k)(m+1)=2km.$$</span> (<span class="math-container">$m$</span> is called the quasi-Euler prime of <span class="math-container">$n$</span>.) Note that we define <span class="math-container">$\sigma(m) := m + 1$</span> even when <span class="math-container">$m$</span> is composite (that is, we <em>pretend</em> that <span class="math-container">$m$</span> is prime).</p>
<p>Notice that the lone Descartes number that is known is
<span class="math-container">$$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$</span></p>
<p>In particular, note that:</p>
<p><strong>(1)</strong> <span class="math-container">$k$</span> is a square.</p>
<p><strong>(2)</strong> <span class="math-container">$\sigma(k)/m = 2k - \sigma(k)$</span></p>
<p><strong>(3)</strong> <span class="math-container">$m \equiv 1 \pmod 4$</span></p>
<p>I want to prove that it must necessarily be the case that <span class="math-container">$m < k$</span>, for a Descartes number <span class="math-container">$n = km$</span>.</p>
<p><strong>Lemma 1</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span>, then <span class="math-container">$k \neq m$</span>.</p>
<p>To this end, suppose that <span class="math-container">$m = k$</span>. Then we have
<span class="math-container">$$\frac{\sigma(k)}{k} = \frac{2m}{m + 1} = \frac{2k}{k + 1},$$</span>
from which it follows that
<span class="math-container">$$\sigma(k) = \frac{2k^2}{k + 1} = \frac{2k^2 - 2}{k + 1} + \frac{2}{k + 1} = \frac{2(k - 1)(k + 1)}{k + 1} + \frac{2}{k + 1} = 2(k - 1) + \frac{2}{k + 1}.$$</span></p>
<p>Since <span class="math-container">$\sigma(k)$</span> and <span class="math-container">$2(k - 1)$</span> are integers, it follows that <span class="math-container">$2/(k+1)$</span> is also an integer, which means that <span class="math-container">$(k + 1) \mid 2$</span>. This implies that <span class="math-container">$k + 1 \leq 2$</span>, from which wet get <span class="math-container">$m = k \leq 1$</span>. This last inequality contradicts the condition <span class="math-container">$1 < k$</span>, <span class="math-container">$1 < m$</span>.</p>
<p><strong>Lemma 2</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, then
<span class="math-container">$$\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}.$$</span></p>
<p>Suppose to the contrary that <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m,k)=1$</span>, and that <span class="math-container">$\sigma(m)/k = \sigma(k)/m$</span>. Then it follows that
<span class="math-container">$$\frac{\sigma(m)}{k} = \frac{\sigma(k)}{m} = r \in \mathbb{N},$$</span>
from which we obtain
<span class="math-container">$$\frac{\sigma(m)}{k}\cdot\frac{\sigma(k)}{m} = r^2 \in \mathbb{N},$$</span>
contradicting
<span class="math-container">$$\sigma(k)\sigma(m) = \sigma(k)(m+1) = 2km,$$</span>
since the last two equations imply that <span class="math-container">$r^2 = 2$</span>.</p>
<p><strong>Lemma 3</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then we have</p>
<p><strong>(a)</strong> <span class="math-container">$\sigma(m) \neq \sigma(k)$</span></p>
<p><strong>(b)</strong> <span class="math-container">$\sigma(m) \neq k$</span></p>
<p><strong>(c)</strong> <span class="math-container">$\sigma(k) \neq m$</span></p>
<p>Proof of <strong>(a)</strong>: Suppose that <span class="math-container">$\sigma(k) = \sigma(m) = m + 1 \equiv 2 \pmod 4$</span>. This contradicts the fact that <span class="math-container">$k$</span> is a square, since then <span class="math-container">$\sigma(k) \equiv 1 \pmod 2$</span>.</p>
<p>Proof of <strong>(b)</strong>: Suppose that <span class="math-container">$\sigma(m) = k$</span>. Then the even number <span class="math-container">$m + 1 = \sigma(m)$</span> is equal to the odd number <span class="math-container">$k$</span>, which is a clear contradiction.</p>
<p>Proof of <strong>(c)</strong>: Suppose to the contrary that <span class="math-container">$\sigma(k) = m$</span>. Then we obtain the estimate
<span class="math-container">$$\frac{\sigma(k)}{m} + \frac{\sigma(m)}{k} = 1 + 2 = 3,$$</span>
which contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Lemma 4</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k)=1$</span>, then the following biconditionals hold:
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}$$</span></p>
<p>We consider three different cases:</p>
<p><strong>Case (1)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} = \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (1) is equivalent to <span class="math-container">$\sigma(m) = \sigma(k)$</span> (which is ruled out by Lemma 3 (a)) or <span class="math-container">$k = m$</span> (which is ruled out by Lemma 1).</p>
<p><strong>Case (2)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$</span></p>
<p>Case (2) implies the estimate
<span class="math-container">$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{9 + 1}{9} + 2 = \frac{28}{9},$$</span>
which again contradicts the known quantity
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$</span></p>
<p><strong>Case (3)</strong>:
<span class="math-container">$$\frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{\sigma(m)}{k} + \frac{\sigma(k)}{m}$$</span></p>
<p>Case (3) is equivalent to the truth of the biconditional <span class="math-container">$m < k \iff \sigma(m) < \sigma(k)$</span> (by virtue of Lemma 1, Lemma 2, and Lemma 3), which in turn is equivalent to the truth of the biconditional
<span class="math-container">$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}.$$</span></p>
<p>By Lemma 4, we have the following possibilities:</p>
<p><strong>(A)</strong> <span class="math-container">$k < \sigma(k) < m < \sigma(m)$</span></p>
<p><strong>(B)</strong> <span class="math-container">$m < \sigma(m) < k < \sigma(k)$</span></p>
<p>Note that Case <strong>(A)</strong> implies that
<span class="math-container">$$\frac{\sigma(k)}{m} = 2k - \sigma(k) < 1$$</span>
forcing <span class="math-container">$2k - \sigma(k) = 0$</span> (i.e. <span class="math-container">$k$</span> must be perfect). This contradicts the fact that <span class="math-container">$k$</span> is a square.</p>
<p>Hence we necessarily have Case <strong>(B)</strong>, and a proof for the following theorem:</p>
<p><strong>THEOREM</strong> If <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span> and <span class="math-container">$\gcd(m, k) = 1$</span>, then <span class="math-container">$k$</span> is <em>not</em> an odd almost perfect number.</p>
<p><strong>QUESTIONS</strong></p>
<p><em>(I)</em> Can we remove the reliance of the proof on the condition
<span class="math-container">$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002?$$</span></p>
<p><em>(II)</em> To what extent can we relax the condition <span class="math-container">$\gcd(m, k)=1$</span> in the <strong>THEOREM</strong>?</p>
| Jose Arnaldo Bebita Dris | 28,816 | <p>(This is not an answer - just some comments that are too long to fit in the appropriate section.)</p>
<p>Note that the inequality <span class="math-container">$m < k$</span> is equivalent to the assertion "<span class="math-container">$k$</span> is not an odd almost perfect number", where <span class="math-container">$n = km$</span> is a Descartes number with quasi-Euler prime <span class="math-container">$m$</span>.</p>
<p>This is proved in the paper "The Non-Euler Part of a Spoof Odd Perfect Number is not Almost Perfect", which is joint work between Dris and Tejada, and published in the <a href="https://ijntindia.puzl.com/archives" rel="nofollow noreferrer">Indian Journal in Number Theory</a>.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.